Free math worksheets

- First determine which input of \(f^{\prime}\) corresponds to \(x=2\). - The chain rule combines an outer derivative value with an inner derivative value. - All needed values are given; no explicit formulas for \(f\) or \(g\) are required.

1. By the chain rule, \(h^{\prime}(x)=f^{\prime}(g(x))g^{\prime}(x)\). 2. At \(x=2\), the inner value is \(g(2)=5\), so \(f^{\prime}(g(2))=f^{\prime}(5)=-4\). 3. Therefore, \(h^{\prime}(2)=(-4)(3)=-12\).

\(-12\)

- Identify the outer power and the inner linear expression separately. - After differentiating the outer function, check whether its input also changes with \(x\). - Compare the student's expression with the full chain-rule structure.

1. The outer power contributes \(4(2x+1)^3\). 2. The inner function \(2x+1\) has derivative \(2\), so the chain rule requires an additional factor of \(2\). 3. Thus, \(p^{\prime}(x)=8(2x+1)^3\).

The missing factor is \(2\), and \(p^{\prime}(x)=8(2x+1)^3\).

- Recall the product rule for \((uv)'(x)\). - Differentiate \(u\) and \(v\) separately; the chain rule can be used for \(v\). - Compute both \((uv)'(x)\) and \(u'(x)v'(x)\). - Compare the expressions over the full domain \(x \ne 1\).

1. Differentiate the factors: \(u'(x) = 1\) and, by the chain rule, \(v'(x) = \frac{1}{(1-x)^2}\). 2. Use the product rule: \((uv)'(x) = u'(x)v(x) + u(x)v'(x)\) \(= \frac{1}{1-x} + \frac{x}{(1-x)^2}\) \(= \frac{1-x+x}{(1-x)^2} = \frac{1}{(1-x)^2}\). 3. The product of the separate derivatives is \(u'(x)v'(x) = \frac{1}{(1-x)^2}\). Thus the two expressions are equal for every \(x \ne 1\), so the claim is true for this specific pair of functions, even though it is not a general derivative rule.

The claim is true for these functions because both sides equal \(\frac{1}{(1-x)^2}\) for every \(x \ne 1\).

- Identify the inner function and the outer power function. - Use the chain rule: outer derivative times inner derivative. - Include the constant factor \(\frac{1}{4}\) when differentiating the outer function.

1. Let the inner function be \(u(z)=2z^3+5\). Then \(u'(z)=6z^2\). 2. The outer function is \(\frac{1}{4}u^4\), whose derivative with respect to \(u\) is \(u^3\). 3. Apply the chain rule: \(h'(z)=(2z^3+5)^3(6z^2)\) \(=6z^2(2z^3+5)^3\).

\(h'(z)=6z^2(2z^3+5)^3\)

- Differentiate the two terms separately. - The sine term is a composite function. - What is the derivative of the expression inside the sine? - Multiply by the inner derivative when applying the chain rule.

1. Differentiate the two terms separately. 2. For the sine term, the inner function is \(u(t)=4t^2-\pi\), with \(u'(t)=8t\). By the chain rule, \(\frac{d}{dt}[2\sin(4t^2-\pi)]=2\cos(4t^2-\pi)(8t)\) \(=16t\cos(4t^2-\pi)\). 3. Also, \(\frac{d}{dt}(t^2)=2t\). Therefore, \(k'(t)=16t\cos(4t^2-\pi)+2t\).

\(k'(t)=16t\cos(4t^2-\pi)+2t\)

- Count the layers of composition from the outside inward. - Each layer should contribute a factor to the derivative. - What is the innermost function, and what is its derivative? - Rewriting the function with parentheses can make the layers clearer.

1. The function has three layers: the square, the sine function, and the inner linear function \(5x\). 2. The student differentiated the square and the sine but did not multiply by the derivative of \(5x\). 3. Applying the chain rule through all layers gives \(f'(x)=2\sin(5x)\cos(5x)\cdot 5\) \(=10\sin(5x)\cos(5x)\).

The derivative of the innermost function \(5x\) was omitted. The correct derivative is \(f'(x)=10\sin(5x)\cos(5x)\).

- Identify the inner and outer functions. - Use the chain rule for each composite function. - Multiply by the derivative of the inner function. - Find the general derivative before substituting the given input.

1. For a), apply the chain rule: \(f'(x)=4(2x-5)^3\cdot 2=8(2x-5)^3\). Therefore, \(f'(3)=8 \cdot (6-5)^3=8\). 2. For b), \(g'(t)=\frac{1}{2\sqrt{t^2+7}}(2t)=\frac{t}{\sqrt{t^2+7}}\). Therefore, \(g'(3)=\frac{3}{\sqrt{16}}=\frac{3}{4}\).

a) \(f'(3)=8\) b) \(g'(3)=\frac{3}{4}\)

- Recall the derivative of an exponential function and the basic trigonometric functions. - Rewrite \(\sin^2(x)\) as \((\sin(x))^2\) to see the composition. - You may simplify \(2\sin(x)\cos(x)\) using a double-angle identity.

1. For a), apply the chain rule: \(f'(x)=e^{x^2-1}(2x)\). Therefore, \(f'(1)=e^0 \cdot 2=2\). 2. For b), write \(g(x)=(\sin(x))^2\). Then \(g'(x)=2\sin(x)\cos(x)\). At \(x=\frac{\pi}{4}\), \(g'\left(\frac{\pi}{4}\right)=2 \cdot \left(\frac{\sqrt{2}}{2}\right) \cdot \left(\frac{\sqrt{2}}{2}\right)=1\).

a) \(f'(1)=2\) b) \(g'\left(\frac{\pi}{4}\right)=1\)

- The operation performed last is usually the outer function. - Identify the expression inside the outer power. - When differentiating the outer function, temporarily keep the inner input unchanged.

1. Choose \(u(x)=x^5\) and \(v(x)=2x^3+7\). 2. Their derivatives are \(u'(x)=5x^4\) and \(v'(x)=6x^2\). 3. Substitute \(v(x)\) into \(u'\): \(u'(v(x))=5(2x^3+7)^4\). 4. By the chain rule, \(f'(x)=u'(v(x))v'(x)\) \(=5(2x^3+7)^4(6x^2)\) \(=30x^2(2x^3+7)^4\).

a) \(u(x)=x^5\); \(v(x)=2x^3+7\) b) \(u'(x)=5x^4\); \(v'(x)=6x^2\) c) \(u'(v(x))=5(2x^3+7)^4\); \(f'(x)=30x^2(2x^3+7)^4\)

- Identify the inner and outer functions. - Use the rule “outer derivative times inner derivative.” - Differentiate the power while keeping the inner expression unchanged. - What is the derivative of the expression inside the parentheses?

1. Let the inner function be \(v(x)=2x^3-4\), so \(v'(x)=6x^2\). 2. The derivative of the outer function \(\frac{1}{2}v^5\) is \(\frac{5}{2}v^4\). 3. Apply the chain rule: \(f'(x)=\frac{5}{2}(2x^3-4)^4(6x^2)\) \(=15x^2(2x^3-4)^4\).

\(f'(x)=15x^2(2x^3-4)^4\)

- Identify which function is applied first and which is applied second. - Recall the chain rule for a composition. - Differentiate the outer and inner functions separately. - Multiply the outer derivative evaluated at the inner function by the inner derivative.

1. Let the outer function be \(u(v)=e^v\) and the inner function be \(v(x)=\sin(x)\). 2. Their derivatives are \(u'(v)=e^v\) and \(v'(x)=\cos(x)\). 3. By the chain rule, \(f'(x)=u'(v(x))v'(x)\) \(=e^{\sin(x)}\cos(x)\), which is the stated expression.

With \(u(v)=e^v\) and \(v(x)=\sin(x)\), the chain rule gives \(f'(x)=e^{\sin(x)}\cos(x)\).

- Identify the inner and outer functions. - Recall the derivatives of the natural logarithm and cosine. - Pay attention to the sign when differentiating the inner function. - Use a trigonometric identity to simplify the quotient.

1. Let the outer function be \(u(v)=\ln(v)\) and the inner function be \(v(x)=\cos(x)\). 2. Their derivatives are \(u'(v)=\frac{1}{v}\) and \(v'(x)=-\sin(x)\). 3. By the chain rule, \(g'(x)=u'(v(x))v'(x)\) \(=\frac{1}{\cos(x)}(-\sin(x))\) \(=-\frac{\sin(x)}{\cos(x)}=-\tan(x)\).

The chain rule gives \(g'(x)=\frac{1}{\cos(x)}(-\sin(x))=-\tan(x)\).

- Identify the operation performed last; it is usually the outer function. - Look for the expression inside the parentheses or under the radical. - Use the chain rule: derivative of the outer function times derivative of the inner function. - Rewriting a square root as a fractional power may help.

1. For a), choose \(u(t)=t^6\) and \(v(x)=2x^2-1\). Then \(u'(t)=6t^5\) and \(v'(x)=4x\). By the chain rule, \(f'(x)=6(2x^2-1)^5(4x)=24x(2x^2-1)^5\). 2. For b), choose \(u(t)=4\sqrt{t}\) and \(v(x)=x^3+2\). Then \(u'(t)=\frac{2}{\sqrt{t}}\) and \(v'(x)=3x^2\). By the chain rule, \(f'(x)=\frac{2}{\sqrt{x^3+2}}(3x^2)=\frac{6x^2}{\sqrt{x^3+2}}\). This derivative is defined for \(x>-\sqrt[3]{2}\).

a) \(u(t)=t^6\), \(v(x)=2x^2-1\); \(f'(x)=24x(2x^2-1)^5\) b) \(u(t)=4\sqrt{t}\), \(v(x)=x^3+2\); \(f'(x)=\frac{6x^2}{\sqrt{x^3+2}}\) for \(x>-\sqrt[3]{2}\)

- Identify the expression used as the input to the trigonometric function or power. - In a), keep the entire derivative of the inner function together. - Ask what operation is performed first and what operation is performed afterward. - Remember the negative sign in the derivative of cosine.

1. For a), choose \(u(t)=\cos(t)\) and \(v(x)=2x^4-x\). Since \(u'(t)=-\sin(t)\) and \(v'(x)=8x^3-1\), the chain rule gives \(f'(x)=-\sin(2x^4-x)(8x^3-1)\) \(=(1-8x^3)\sin(2x^4-x)\). 2. For b), choose \(u(t)=t^2\) and \(v(x)=\sin(x)-5\). Since \(u'(t)=2t\) and \(v'(x)=\cos(x)\), the chain rule gives \(f'(x)=2(\sin(x)-5)\cos(x)\).

a) \(u(t)=\cos(t)\), \(v(x)=2x^4-x\); \(f'(x)=(1-8x^3)\sin(2x^4-x)\) b) \(u(t)=t^2\), \(v(x)=\sin(x)-5\); \(f'(x)=2(\sin(x)-5)\cos(x)\)

- Identify the expression inside each exponent. - Differentiate the outer exponential function, then multiply by the derivative of the exponent. - Keep constant factors. - Rewrite a reciprocal exponential using a negative exponent.

1. For \(f(x)=e^{4x-2}\), the inner derivative is \(4\), so \(f'(x)=4e^{4x-2}\). 2. For \(g(x)=5e^{-0.2x}\), multiply by the inner derivative \(-0.2\): \(g'(x)=5\cdot(-0.2)e^{-0.2x}=-e^{-0.2x}\). 3. For \(h(x)=e^{x^2+3}\), the inner derivative is \(2x\), so \(h'(x)=2xe^{x^2+3}\). 4. Rewrite \(k(x)=e^{-2x}\). Then \(k'(x)=-2e^{-2x}=-\frac{2}{e^{2x}}\).

a) \(f'(x)=4e^{4x-2}\) b) \(g'(x)=-e^{-0.2x}\) c) \(h'(x)=2xe^{x^2+3}\) d) \(k'(x)=-2e^{-2x}=-\frac{2}{e^{2x}}\)

- Identify and differentiate the expression in each exponent. - Multiply the outer derivative by the inner derivative. - Rewrite a reciprocal exponential using a negative exponent. - Track negative signs carefully.

1. For \(f(x)=0.2e^{5-x^2}\), the inner derivative is \(-2x\). Thus, \(f'(x)=0.2\cdot(-2x)e^{5-x^2}=-0.4xe^{5-x^2}\). 2. For \(f(x)=e^{\frac{1}{2}x^2-x}\), the inner derivative is \(x-1\). Thus, \(f'(x)=(x-1)e^{\frac{1}{2}x^2-x}\). 3. Rewrite \(f(x)=-3e^{-4x-1}\). Then \(f'(x)=12e^{-4x-1}=\frac{12}{e^{4x+1}}\). 4. For \(f(x)=4e^{1-2x^3}\), the inner derivative is \(-6x^2\). Thus, \(f'(x)=-24x^2e^{1-2x^3}\).

a) \(f'(x)=-0.4xe^{5-x^2}\) b) \(f'(x)=(x-1)e^{\frac{1}{2}x^2-x}\) c) \(f'(x)=12e^{-4x-1}=\frac{12}{e^{4x+1}}\) d) \(f'(x)=-24x^2e^{1-2x^3}\)

- Identify the inner and outer functions in each term. - Multiply by the derivative of the inner function when applying the chain rule. - Remember the negative sign in the derivative of cosine.

1. For \(f(x)=\sin(4x)\), differentiate the outer sine function and multiply by the derivative of \(4x\): \(f'(x)=\cos(4x)\cdot 4=4\cos(4x)\). 2. For \(f(x)=5\cos(2x-1)\), \(f'(x)=5[-\sin(2x-1)]\cdot 2=-10\sin(2x-1)\). 3. Differentiate each term: \(\frac{d}{dx}\sin(0.5x)=0.5\cos(0.5x)\) and \(\frac{d}{dx}\cos(x^2)=-2x\sin(x^2)\). Therefore, \(f'(x)=0.5\cos(0.5x)-2x\sin(x^2)\).

1) \(f'(x)=4\cos(4x)\) 2) \(f'(x)=-10\sin(2x-1)\) 3) \(f'(x)=0.5\cos(0.5x)-2x\sin(x^2)\)

- First differentiate both functions using the chain rule. - Recall the derivatives of sine and cosine, including the sign. - Solve the Pythagorean identity for \(\sin^2(x)\). - Consider how adding a constant affects a derivative.

1. Apply the chain rule to \(u(x)=(\sin(x))^2\): \(u'(x)=2\sin(x)\cos(x)=\sin(2x)\). 2. Apply the chain rule to \(v(x)=-(\cos(x))^2\): \(v'(x)=-2\cos(x)(-\sin(x))=2\sin(x)\cos(x)=\sin(2x)\). 3. Thus, \(u'(x)=v'(x)\). 4. The Pythagorean identity gives \(\sin^2(x)=1-\cos^2(x)\), so \(u(x)=v(x)+1\). Functions that differ by a constant have the same derivative.

a) \(u'(x)=\sin(2x)\) and \(v'(x)=\sin(2x)\) b) The derivatives are equal because \(u(x)=v(x)+1\), so the functions differ only by a constant.

- Identify the two factors in the function. - Use the chain rule when differentiating \(e^{2x}\). - Factor out the common exponential term at the end.

1. Let \(u(x) = 3x^2 - 5x\) and \(v(x) = e^{2x}\). 2. Then \(u'(x) = 6x - 5\) and, by the chain rule, \(v'(x) = 2e^{2x}\). 3. Apply the product rule: \(f'(x) = (6x-5)e^{2x} + (3x^2-5x)(2e^{2x})\). 4. Factor out \(e^{2x}\) and combine like terms: \(f'(x) = (6x^2 - 4x - 5)e^{2x}\).

\(f'(x) = (6x^2 - 4x - 5)e^{2x}\)

- Rewrite reciprocal exponentials using negative exponents. - Simplify the square root before differentiating. - Use the chain rule for each exponential function. - Treat \(k\) as a constant.

1. Rewrite the first function as \(f(x)=e^{4x}-3e^{-x}\). Then \(f'(x)=4e^{4x}+3e^{-x}\). 2. Since \(e^{6x}>0\), \(\sqrt{e^{6x}}=e^{3x}\). Therefore, \(f'(x)=3e^{3x}\). 3. Treat \(k\) as a constant. The inner derivative of \(kx^2\) is \(2kx\), so \(f'(x)=2kxe^{kx^2}\).

a) \(f'(x)=4e^{4x}+3e^{-x}\) b) \(f'(x)=3e^{3x}\) c) \(f'(x)=2kxe^{kx^2}\)

- Determine whether the radical or denominator can ever be zero. - Rewrite a radical denominator using a negative rational exponent. - Apply the chain rule to the composite power. - Multiply by the derivative of \(t^2+4\).

1. Since \(t^2+4\ge4\) for every real \(t\), the radical is positive and the denominator is never zero. Thus, \(D=\mathbb{R}\). 2. Rewrite \(g(t)=10(t^2+4)^{-\frac{1}{2}}\). 3. Apply the chain rule: \(g'(t)=10\left(-\frac{1}{2}\right)(t^2+4)^{-\frac{3}{2}}(2t)=-\frac{10t}{(t^2+4)^{\frac{3}{2}}}\).

\(D=\mathbb{R}\); \(g'(t)=-\frac{10t}{(t^2+4)^{\frac{3}{2}}}\)

- The slope at an input is the derivative value there. - Use the quotient property of logarithms before differentiating. - Include the inner derivative of \(2x+1\).

1. Rewrite the function: \(g(x)=\ln4-\ln(2x+1)\). 2. Differentiate using the chain rule: \(g'(x)=-\frac{2}{2x+1}\). 3. Evaluate at \(x=1.5\): \(g'(1.5)=-\frac{2}{2\cdot1.5+1}=-\frac{1}{2}\).

\(-\frac{1}{2}\)

- Identify the inner function inside the logarithm. - Differentiate the outer and inner functions. - Combine them using the chain rule. - Check whether the denominator can be zero when \(x>0\).

1. Apply the chain rule with inner function \(v(x)=x^2+4x\): \(f'(x)=\frac{v'(x)}{v(x)}=\frac{2x+4}{x^2+4x}\). 2. For every \(x>0\), the denominator \(x^2+4x=x(x+4)\) is positive. Therefore, the derivative is defined throughout the stated interval.

\(f'(x)=\frac{2x+4}{x^2+4x}\), with domain \((0, \infty)\)

- The natural logarithm equals zero when its input equals \(1\). - Solve the resulting linear equation. - Apply the chain rule to the logarithm.

1. Solve \(\ln(2x+e)=0\). Then \(2x+e=1\), so \(x=\frac{1-e}{2}\). 2. Apply the chain rule: \(g'(x)=\frac{2}{2x+e}\).

Zero: \(x=\frac{1-e}{2}\) \(g'(x)=\frac{2}{2x+e}\)

- Identify the inner and outer functions for the chain-rule method. - Use the square of a difference to expand the expression. - Expand the chain-rule result before comparing the two answers.

1. Using the chain rule, \(f'(x)=2(2x^3-5)(6x^2)\) \(=12x^2(2x^3-5)=24x^5-60x^2\). 2. Expanding first gives \(f(x)=4x^6-20x^3+25\). Differentiating term by term, \(f'(x)=24x^5-60x^2\). Both methods give the same derivative.

\(f'(x)=24x^5-60x^2\)

- An affine function has the form \(y=px+q\), including the case \(p=0\). - Substitute the entire formula for the inner function into the outer function. - Use the chain rule: outer derivative times inner derivative. - Check what happens when one of the slopes is zero.

1. For a), substitute \(k(x)\) into \(g\): \(L(x)=m(ax+b)+c=max+mb+c\). This is affine, with slope \(ma\). 2. For b), \(h(x)=(mx+c)^2\). By the chain rule, \(h'(x)=2(mx+c)m=2m^2x+2mc\). 3. The derivative of the outer quadratic function is affine. Composing that derivative with the inner affine function produces another affine function, as shown in a). Multiplying by the constant inner derivative \(m\) preserves this form. Therefore, \(h'\) is affine; when \(m=0\), it is the zero function.

a) \(L(x)=max+mb+c\), so its slope is \(ma\). b) \(h'(x)=2m(mx+c)=2m^2x+2mc\) c) The derivative is a constant multiple of a composition of two affine functions, so it is affine (possibly constant).

- Identify the inner and outer functions. - Apply the rule “outer derivative times inner derivative.” - Pay close attention to signs, especially for cosine and the decreasing linear expression under the square root.

1. For a), \(f'(x)=\cos(x^2+3)(2x)=2x\cos(x^2+3)\). 2. For b), \(f'(x)=\frac{1}{2\sqrt{10-2x}}\cdot(-2)=-\frac{1}{\sqrt{10-2x}}\). This derivative is defined for \(x<5\). 3. For c), \(f'(x)=0.5[-\sin(4x)]\cdot 4=-2\sin(4x)\).

a) \(f'(x)=2x\cos(x^2+3)\) b) \(f'(x)=-\frac{1}{\sqrt{10-2x}}\) for \(x<5\) c) \(f'(x)=-2\sin(4x)\)

- Use the point condition independently of the derivative condition. - Differentiate the exponential factor with its inner function in mind. - Evaluate the derivative only after simplifying the product-rule result.

1. Substitute the point: \((2+a)e^{-2}=5e^{-2}\). Dividing by \(e^{-2}\) gives \(a=3\). 2. By the product and chain rules, \(g_a'(x)=(1-x-a)e^{-x}\). 3. Thus \(g_a'(0)=1-a\). Setting \(1-a=-2\) gives \(a=3\).

1. \(a=3\) 2. \(a=3\)

- A natural logarithm equals \(0\) when its input equals \(1\). - Use the chain rule for \(\ln(u(x))\). - Simplify the derivative after applying the rule.

1. For the \(x\)-intercept, solve \(\ln(6-2x)=0\). Then \(6-2x=1\), so \(x=\frac{5}{2}\). The intercept is \(\left(\frac{5}{2}, 0\right)\). 2. By the chain rule, \(f'(x)=\frac{-2}{6-2x}=\frac{1}{x-3}\).

\(x\)-intercept: \(\left(\frac{5}{2}, 0\right)\); \(f'(x)=\frac{1}{x-3}\)

- Set the function equal to zero and isolate the exponential expression. - Apply the natural logarithm. - Differentiate the exponential function using the chain rule.

1. Set \(h(x)=0\): \(3-e^{x-1}=0\), so \(e^{x-1}=3\). 2. Taking natural logarithms gives \(x-1=\ln(3)\), so the \(x\)-intercept is \((1+\ln(3), 0)\). 3. Differentiate: \(h'(x)=-e^{x-1}\).

\(x\)-intercept: \((1+\ln(3), 0)\); \(h'(x)=-e^{x-1}\)

- Use the chain rule when differentiating \(\sin(2x)\). - Evaluate both the function and derivative at the given input. - Use point-slope form for the tangent line. - For a negative slope, choose the inclination angle in the second quadrant.

1. Find the point of tangency: \(f\left(\frac{\pi}{3}\right)=2\sin\left(\frac{2\pi}{3}\right)+\cos\left(\frac{\pi}{3}\right)=\sqrt{3}+\frac{1}{2}\). 2. Differentiate using the chain rule: \(f'(x)=4\cos(2x)-\sin x\). The tangent slope is \(m=f'\left(\frac{\pi}{3}\right)=-2-\frac{\sqrt{3}}{2}\). 3. Use point-slope form: \(y-\left(\sqrt{3}+\frac{1}{2}\right)=\left(-2-\frac{\sqrt{3}}{2}\right)\left(x-\frac{\pi}{3}\right)\). Equivalently, \(y=\left(-2-\frac{\sqrt{3}}{2}\right)x+\frac{2\pi}{3}+\frac{\pi\sqrt{3}}{6}+\sqrt{3}+\frac{1}{2}\). 4. The angle of inclination \(\alpha\in[0^\circ, 180^\circ)\) satisfies \(\tan\alpha=m\). Since the slope is negative, \(\alpha\approx 109.23^\circ\).

Tangent line: \(y-\left(\sqrt{3}+\frac{1}{2}\right)=\left(-2-\frac{\sqrt{3}}{2}\right)\left(x-\frac{\pi}{3}\right)\) Angle of inclination: \(\alpha\approx 109.23^\circ\)

- Recall the product rule for \((fg)'(x)\). - Differentiate \(f\) and \(g\) separately. - Factor the common exponential term from both expressions. - Compare the coefficients of the common exponential factor.

1. Differentiate the two functions: \(f'(x) = 2e^{2x}\) and \(g'(x) = ke^{kx}\). 2. Use the product rule: \((fg)'(x) = f'(x)g(x) + f(x)g'(x)\) \(= 2e^{2x}e^{kx} + ke^{2x}e^{kx} = (2+k)e^{(2+k)x}\). 3. The product of the separate derivatives is \(f'(x)g'(x) = 2ke^{(2+k)x}\). 4. Set the expressions equal. Since \(e^{(2+k)x}\) is never zero, \(2+k = 2k\). Thus \(k = 2\).

\(k = 2\)

- The chain rule requires multiplying by the derivative of the inner function. - Differentiate the expression inside the parentheses separately. - Evaluate the derivative at the specified \(x\)-value to find the slope. - Track the sign when raising a negative number to an odd power.

1. The student differentiated the outer fourth power but omitted the derivative of the inner function. 2. The inner function is \(u(x)=2x^3-5\), with \(u'(x)=6x^2\). 3. By the chain rule, \(g'(x)=4(2x^3-5)^3(6x^2)\) \(=24x^2(2x^3-5)^3\). 4. Evaluate at \(x=1\): \(g'(1)=24 \cdot 1^2 \cdot (2-5)^3=24 \cdot (-3)^3=-648\).

The inner derivative was omitted. The correct derivative is \(g'(x)=24x^2(2x^3-5)^3\), and the slope at \(x=1\) is \(-648\).

- The chain rule has the form “outer derivative times inner derivative.” - Keep the original inner expression as the input of the differentiated outer function. - For an exponential function, check whether the original exponent has been preserved. - Look for factors that should have appeared from differentiating the inner function.

1. In a), the inner derivative is missing. Since \(\frac{d}{dx}(x^5-2)=5x^4\), \(f'(x)=4(x^5-2)^3(5x^4)=20x^4(x^5-2)^3\). 2. In b), the derivative of the inner function was incorrectly substituted into the sine argument. The original inner expression remains inside the outer derivative: \(g'(x)=-\sin(3x^2)(6x)=-6x\sin(3x^2)\). 3. In c), the exponent was incorrectly changed. For \(e^{u(x)}\), the exponential factor remains \(e^{u(x)}\): \(h'(x)=e^{4x+1}(4)=4e^{4x+1}\).

a) The inner derivative was omitted. \(f'(x)=20x^4(x^5-2)^3\). b) The argument of the sine function was changed incorrectly. \(g'(x)=-6x\sin(3x^2)\). c) The exponent was changed incorrectly. \(h'(x)=4e^{4x+1}\).

- Recall the derivative of the natural exponential function. - What sign appears when cosine is differentiated? - A slope at a point is found by evaluating the derivative there. - Use the unit-circle values of sine and cosine at \(\frac{\pi}{2}\).

1. Choose \(u(x)=e^x\) and \(v(x)=\cos(x)\). 2. Then \(u'(x)=e^x\), \(v'(x)=-\sin(x)\), and \(u'(v(x))=e^{\cos(x)}\). 3. By the chain rule, \(f'(x)=u'(v(x))v'(x)\) \(=-\sin(x)e^{\cos(x)}\). 4. At \(x=\frac{\pi}{2}\), \(f'\left(\frac{\pi}{2}\right)=-\sin\left(\frac{\pi}{2}\right)e^{\cos(\frac{\pi}{2})}\) \(=-1\cdot e^0=-1\).

a) \(u(x)=e^x\); \(v(x)=\cos(x)\) b) \(u'(x)=e^x\); \(v'(x)=-\sin(x)\); \(u'(v(x))=e^{\cos(x)}\) c) \(f'(x)=-\sin(x)e^{\cos(x)}\) d) \(f'\left(\frac{\pi}{2}\right)=-1\)

- Differentiate the two terms separately. - Rewrite the rational term using a negative exponent. - Identify the inner function in the cosine term. - Watch the negative sign when differentiating cosine. - Apply the chain rule to both terms.

1. Differentiate the first term using the chain rule: \(\frac{d}{dt}[3\cos(2t^2+5)]=-3\sin(2t^2+5)(4t)\) \(=-12t\sin(2t^2+5)\). 2. Rewrite the second term as \(-4(3t-2)^{-1}\). Then \(\frac{d}{dt}[-4(3t-2)^{-1}]=4(3t-2)^{-2}\cdot 3\) \(=\frac{12}{(3t-2)^2}\). 3. Combine the results: \(g'(t)=-12t\sin(2t^2+5)+\frac{12}{(3t-2)^2}\).

\(g'(t)=-12t\sin(2t^2+5)+\frac{12}{(3t-2)^2}\), for \(t\ne\frac{2}{3}\)

- Distinguish the variable from the parameters in each function. - Identify the inner and outer functions. - Watch the sign when differentiating cosine. - Make the power structure explicit before applying the chain rule.

1. For a), apply the chain rule: \(f_k'(x)=4(kx^2-5)^3(2kx)\) \(=8kx(kx^2-5)^3\). 2. For b), \(g_a'(t)=\cos(at-a)(a)\) \(=a\cos(at-a)\). 3. For c), write \(h_b(x)=(\cos(bx))^2\). Applying the chain rule through both layers gives \(h_b'(x)=2\cos(bx)[-b\sin(bx)]\) \(=-2b\sin(bx)\cos(bx)\).

a) \(f_k'(x)=8kx(kx^2-5)^3\) b) \(g_a'(t)=a\cos(at-a)\) c) \(h_b'(x)=-2b\sin(bx)\cos(bx)\)

- Check which variable each function is differentiated with respect to. - Any expression that does not contain the variable acts like a constant. - For composite functions, differentiate the inner function separately. - Simplify an inner expression before differentiating when helpful.

1. For a), the inner function is \(cz^3\), whose derivative is \(3cz^2\). Therefore, \(f_c'(z)=3cz^2\cos(cz^3)\). 2. For b), \((ax)^2=a^2x^2\), whose derivative with respect to \(x\) is \(2a^2x\). Thus, \(g_a'(x)=-2a^2x\sin(a^2x^2)\). 3. For c), \(t\) is the variable and \(\sin(ak^2)\) is a constant factor. Therefore, \(h_{a,k}'(t)=\sin(ak^2)\).

a) \(f_c'(z)=3cz^2\cos(cz^3)\) b) \(g_a'(x)=-2a^2x\sin(a^2x^2)\) c) \(h_{a,k}'(t)=\sin(ak^2)\)

- Use the chain rule for a power of a linear expression. - A horizontal tangent has derivative \(0\). - Convert the given angle to a slope. - Find the y-coordinate after solving for the x-coordinate.

1. Differentiate using the chain rule: \(f'(x)=\frac{1}{2}\cdot 4(0.5x+1)^3\cdot 0.5=(0.5x+1)^3\). 2. At \(x=2\), \(f'(2)=(2)^3=8\). 3. A horizontal tangent has slope \(0\). Solve \((0.5x+1)^3=0\), giving \(x=-2\). Since \(f(-2)=0\), the point is \((-2,0)\). 4. An angle of \(45^\circ\) corresponds to slope \(1\). Solve \((0.5x+1)^3=1\), giving \(x=0\). 5. Since \(f(0)=\frac{1}{2}\), the point is \((0,\frac{1}{2})\).

a) \(8\) b) Yes; the point is \((-2,0)\). c) \((0,\frac{1}{2})\)

- Use the chain rule and track the sign of the inner derivative. - A horizontal tangent has slope \(0\). - Convert \(135^\circ\) to a slope. - A squared equation may have two solutions.

1. Differentiate using the chain rule: \(f'(x)=-(2-x)^2\). 2. At \(x=1\), \(f(1)=\frac{1}{3}\) and \(f'(1)=-1\). The tangent line is \(y-\frac{1}{3}=-(x-1)\), or \(y=-x+\frac{4}{3}\). 3. A horizontal tangent has slope \(0\). Solve \(-(2-x)^2=0\), giving \(x=2\). Since \(f(2)=0\), the point is \((2,0)\). 4. An angle of \(135^\circ\) corresponds to slope \(-1\). Solve \(-(2-x)^2=-1\), so \((2-x)^2=1\). Thus, \(x=1\) or \(x=3\). 5. The corresponding points are \((1,\frac{1}{3})\) and \((3,-\frac{1}{3})\).

a) \(y=-x+\frac{4}{3}\) b) \((2,0)\) c) \((1,\frac{1}{3})\) and \((3,-\frac{1}{3})\)

- For each part, identify the inner and outer functions. - Use the chain rule: differentiate the outer function, then multiply by the derivative of the inner function. - In b), remember how constants behave under differentiation. - Rewriting a reciprocal or square root as a power may help.

1. For a), the inner function is \(e^x\), whose derivative is \(e^x\). Therefore, \(g'(x)=e^x f'(e^x)\). 2. For b), apply the chain rule to the outer function \(u^3\) and the inner function \(f(x)+k\): \(g'(x)=3(f(x)+k)^2f'(x)\). 3. For c), the inner function is \(x^{-1}\), whose derivative is \(-x^{-2}\). Therefore, \(g'(x)=-\frac{1}{x^2}f'\left(\frac{1}{x}\right)\). 4. For d), differentiate the outer function \(\sqrt{u}\) and multiply by the derivative of the inner function: \(g'(x)=\frac{f'(x)}{2\sqrt{f(x)}}\).

a) \(g'(x)=e^x f'(e^x)\) b) \(g'(x)=3(f(x)+k)^2f'(x)\) c) \(g'(x)=-\frac{1}{x^2}f'\left(\frac{1}{x}\right)\) d) \(g'(x)=\frac{f'(x)}{2\sqrt{f(x)}}\)

- Treat \(a\), \(b\), \(c\), and \(d\) as constants. - Identify the inner and outer functions in each composition. - Apply the power rule together with the chain rule. - For \(f(f(x))\), distinguish the derivative of the outer copy of \(f\) from the derivative of the inner copy.

1. For a), the derivative of the inner function \(ax^n+b\) is \(anx^{n-1}\). Thus, \(g'(x)=anx^{n-1}f'(ax^n+b)\). 2. For b), differentiate the outer function \(\frac{1}{n}u^n\), then multiply by \(f'(x)\): \(g'(x)=(f(x))^{n-1}f'(x)\). 3. For c), the outer copy of \(f\) is evaluated at \(f(x)\), while the inner derivative is \(f'(x)\). Therefore, \(g'(x)=f'(f(x))f'(x)\). 4. For d), the constant \(d\) differentiates to zero, the factor \(a\) remains, and the inner linear function has derivative \(b\). Therefore, \(g'(x)=abf'(bx+c)\).

a) \(g'(x)=anx^{n-1}f'(ax^n+b)\) b) \(g'(x)=(f(x))^{n-1}f'(x)\) c) \(g'(x)=f'(f(x))f'(x)\) d) \(g'(x)=abf'(bx+c)\)

- What equation characterizes a function whose graph is symmetric about the y-axis? - Use the chain rule for a function of the form \(f(u(x))\). - What equation characterizes a function whose graph is symmetric about the origin? - When substituting \(-x\) into \(g'(x)\), track the sign of the linear factor carefully.

1. For a), \(g(-x)=f((-x)^2)=f(x^2)=g(x)\). Therefore, \(g\) is even, so its graph is symmetric about the y-axis. 2. For b), apply the chain rule: \(g'(x)=f'(x^2)(2x)=2x f'(x^2)\). 3. For c), \(g'(-x)=2(-x)f'((-x)^2)\) \(=-2x f'(x^2)=-g'(x)\). Therefore, \(g'\) is odd, so its graph is symmetric about the origin.

a) Since \(g(-x)=f((-x)^2)=f(x^2)=g(x)\), the graph of \(g\) is symmetric about the y-axis. b) \(g'(x)=2x f'(x^2)\) c) Since \(g'(-x)=-2x f'(x^2)=-g'(x)\), the graph of \(g'\) is symmetric about the origin.

- Identify the operation performed last; it gives the outermost function. - Work from the outside inward to identify all three layers. - Apply the chain rule once for each layer of the composition. - Multiply the derivative factors in outer-to-inner order.

1. One valid decomposition is \(u(t)=e^t\), \(v(t)=\cos(t)\), and \(w(x)=2x+1\). 2. Their derivatives are \(u'(t)=e^t\), \(v'(t)=-\sin(t)\), and \(w'(x)=2\). 3. Applying the chain rule to the three-function composition gives \(f'(x)=u'(v(w(x)))v'(w(x))w'(x)\). 4. Substitute the functions and derivatives: \(f'(x)=e^{\cos(2x+1)}(-\sin(2x+1))\cdot 2\) \(=-2\sin(2x+1)e^{\cos(2x+1)}\).

a) One valid choice is \(u(t)=e^t\), \(v(t)=\cos(t)\), and \(w(x)=2x+1\). b) \(f'(x)=-2\sin(2x+1)e^{\cos(2x+1)}\)

- Evaluate the function at \(\pi\). - Use the chain rule for \(\cos^2x\). - The derivative value is the tangent slope. - Use the negative reciprocal for the normal slope.

1. Evaluate the function: \(f(\pi)=\cos^2\pi-\sin\pi=1\). Thus, \(P=(\pi, 1)\). 2. Differentiate using the chain rule: \(f'(x)=2\cos x(-\sin x)-\cos x\) \(=-\sin(2x)-\cos x\). Therefore, \(f'(\pi)=1\). 3. The tangent line is \(y-1=x-\pi\), or \(y=x-\pi+1\). 4. The normal slope is \(-1\), so the normal line is \(y-1=-(x-\pi)\), or \(y=-x+\pi+1\).

Tangent line: \(y=x-\pi+1\) Normal line: \(y=-x+\pi+1\)

- Substitute each value of \(a\) and the given value of \(h\) carefully into the difference quotient. - Compare both approximate slopes with \(1\). - Recall the derivative rule for \(a^x\) and consider what happens when \(\ln(a)=1\). - Which differentiation rule applies to \(e^{2x+5}\)?

1. For \(a=2\), \(\frac{2^{0.0001}-1}{0.0001}\approx0.6932\). 2. For \(a=3\), \(\frac{3^{0.0001}-1}{0.0001}\approx1.0987\). 3. In general, \(\frac{d}{dx}a^x=\ln(a)a^x\). Because \(\ln(e)=1\), differentiating an exponential with base \(e\) does not introduce an additional base-dependent factor. Applying the chain rule gives \(f'(x)=2e^{2x+5}\).

a) For base \(2\), the slope is approximately \(0.6932\). For base \(3\), the slope is approximately \(1.0987\). b) Since \(\frac{d}{dx}e^x=e^x\), only the derivative of the exponent contributes an additional factor. Thus, \(f'(x)=2e^{2x+5}\).

- Substitute a convenient x-value to check the proposed intersection. - What graph transformation is represented by replacing \(x\) with \(-x\)? - Compare the signs of the derivatives of the two functions. - Use exponent rules to simplify the product of the function expressions. - Distinguish between behavior as \(x\to\infty\) and as \(x\to-\infty\).

1. Since \(f(0)=e^0=1\) and \(g(0)=e^{-0}=1\), both graphs contain \((0, 1)\). Statement 1 is true. 2. Because \(g(x)=f(-x)\), the graph of \(g\) is the reflection of the graph of \(f\) across the y-axis, not the x-axis. Statement 2 is false. 3. The derivatives are \(f'(x)=e^x>0\) and \(g'(x)=-e^{-x}<0\) for every real \(x\). Their slopes can never be equal. Statement 3 is false. 4. Using exponent rules, \(h(x)=e^xe^{-x}=e^0=1\). The graph of \(h(x)=1\) is horizontal. Statement 4 is true. 5. As \(x\to\infty\), \(e^x\to\infty\), while \(e^{-x}\to0\). Only the graph of \(g\) approaches the x-axis. Statement 5 is false.

1. True 2. False 3. False 4. True 5. False

- Evaluate the function to find the point of tangency. - Differentiate using the chain rule. - Use the derivative value as the tangent slope. - The normal slope is the negative reciprocal of the tangent slope.

1. Find the point: \(f(0)=e^0+e^0=2\), so the point is \((0, 2)\). 2. Differentiate using the chain rule: \(f'(x)=2e^{2x}+e^x\). Thus, the tangent slope is \(m_t=f'(0)=2+1=3\). The tangent line is \(t:y-2=3(x-0)\), or \(t:y=3x+2\). 3. The normal slope is the negative reciprocal: \(m_n=-\frac{1}{3}\). Therefore, \(n:y-2=-\frac{1}{3}(x-0)\), or \(n:y=-\frac{1}{3}x+2\).

\(t:y=3x+2\) \(n:y=-\frac{1}{3}x+2\)

- Translate the point and slope information into two equations. - Use the chain rule for \(e^{2x}\). - Solve the resulting system of two linear equations.

1. The point condition gives \(g_{r,s}(0)=r+s=1\). 2. Differentiate using the chain rule: \(g_{r,s}'(x)=2re^{2x}+se^x\). 3. The slope condition gives \(g_{r,s}'(0)=2r+s=5\). 4. Subtract \(r+s=1\) from \(2r+s=5\) to obtain \(r=4\). 5. Substitute into \(r+s=1\): \(s=-3\).

\(r=4\) and \(s=-3\)

- Use the chain rule for \(e^{ax}\). - A point on the x-axis has y-coordinate \(0\). - Write the tangent line in terms of \(a\). - Substitute the given x-intercept into the tangent equation.

1. Apply the chain rule: \(g_a'(x)=2ae^{ax}\). 2. At \(x=0\), \(g_a(0)=3\) and \(g_a'(0)=2a\). 3. The tangent line is \(y=2ax+3\). 4. It must pass through \((-1, 0)\), so \(0=-2a+3\). 5. Therefore, \(a=\frac{3}{2}\).

a) \(g_a'(x)=2ae^{ax}\) b) \(a=\frac{3}{2}\)

- Find both the function value and slope at \(x=1\). - Simplify the tangent equation before applying the conditions. - Positive slope determines the sign of \(a\). - Set the tangent equation equal to \(0\) to find its x-intercept. - Translate the x-intercept condition into an inequality for \(a\).

1. At \(x=1\), \(g_a(1)=a+2\). 2. Differentiate: \(g_a'(x)=ae^{x-1}\), so the tangent slope at \(x=1\) is \(a\). 3. The tangent line is \(y-(a+2)=a(x-1)\), which simplifies to \(y=ax+2\). 4. Positive slope requires \(a>0\). 5. The x-intercept is \(x=-\frac{2}{a}\). Require \(-\frac{2}{a}>-1\). 6. Since \(a>0\), multiplication by \(a\) preserves the inequality: \(-2>-a\), so \(a>2\).

a) \(y=ax+2\) b) \(a>2\)

- Use the product rule. - Use the chain rule when differentiating the exponential factor. - Factor out the common exponential factor after differentiating. - Combine like terms inside the remaining parentheses.

1. Let \(u(x)=4-x^2\) and \(v(x)=e^{-0.5x}\). Then \(u'(x)=-2x\) and, by the chain rule, \(v'(x)=-0.5e^{-0.5x}\). 2. Apply the product rule: \(f'(x)=(-2x)e^{-0.5x}-0.5(4-x^2)e^{-0.5x}\). 3. Factor out \(e^{-0.5x}\) and simplify: \(f'(x)=(-2x-2+0.5x^2)e^{-0.5x}\) \(=(0.5x^2-2x-2)e^{-0.5x}\).

\(f'(x)=(0.5x^2-2x-2)e^{-0.5x}\)

- Identify the nested functions in the exponent. - Apply the chain rule to \(e^{u(x)}\). - The exponent is itself a composition, so apply the chain rule again. - Recall the derivative of \(\cos(ax)\).

1. Let \(u(x)=\cos(2x)\). Applying the chain rule to the exponential function gives \(g'(x)=\frac{1}{2}e^{\cos(2x)}u'(x)\). 2. Apply the chain rule again to the cosine function: \(u'(x)=-\sin(2x)\cdot 2=-2\sin(2x)\). 3. Substitute and simplify: \(g'(x)=\frac{1}{2}e^{\cos(2x)}(-2\sin(2x))\) \(=-\sin(2x)e^{\cos(2x)}\).

\(g'(x)=-\sin(2x)e^{\cos(2x)}\)

- A horizontal tangent has slope \(0\). - Use both the product rule and the chain rule. - An exponential factor is never zero. - Compare exponential decay or growth with the linear factor.

1. Differentiate using the product and chain rules: \(f'(x)=-2e^{0.5x}+(4-2x)(0.5)e^{0.5x}=-xe^{0.5x}\). 2. A horizontal tangent requires \(f'(x)=0\). Since \(e^{0.5x}>0\), the only solution is \(x=0\). 3. Evaluate the function: \(f(0)=4\), so the point is \((0,4)\). 4. As \(x\to-\infty\), the exponential decay dominates the linear factor, so \(f(x)\to0\). As \(x\to\infty\), the linear factor approaches \(-\infty\) while the exponential factor grows without bound, so \(f(x)\to-\infty\).

Horizontal-tangent point: \((0,4)\). \(\lim_{x\to-\infty}f(x)=0\), and \(\lim_{x\to\infty}f(x)=-\infty\).

- The derivative gives the graph’s slope. - Combine the product and chain rules. - Evaluate the derivative at \(x=1\). - Divide by the exponential factor, which is never zero.

1. Apply the product and chain rules: \(g_a'(x)=(ax^2+2x)e^{ax}\). 2. The slope condition is \(g_a'(1)=e^a\), so \((a+2)e^a=e^a\). 3. Since \(e^a>0\), divide by \(e^a\): \(a+2=1\). 4. Therefore, \(a=-1\).

\(a=-1\)

- Convert the point information into an equation. - Differentiate to convert the slope information into a second equation. - Two unknown parameters require two independent equations. - Substitute the repeated expression from one equation into the other.

1. The point condition gives \(ae^b=e^2\). 2. Apply the chain rule: \(f_{a,b}'(x)=2abxe^{bx^2}\). 3. The slope condition at \(x=1\) gives \(2abe^b=4e^2\). 4. Substitute \(ae^b=e^2\): \(2be^2=4e^2\), so \(b=2\). 5. Then \(ae^2=e^2\), so \(a=1\).

\(a=1\) and \(b=2\)

- Compare the outputs as the x-values increase. - For consecutive x-values, how is the ratio of the outputs related to the base? - Substitute one point after finding \(a\) to determine \(b\). - Use \(a=e^{\ln(a)}\) to rewrite the function. - Apply the chain rule to differentiate \(e^{kx}\).

1. The function value increases from \(45\) at \(x=1\) to \(135\) at \(x=2\), so the function represents exponential growth. 2. Increasing \(x\) by \(1\) multiplies the output by \(a\). Therefore, \(a=\frac{135}{45}=3\). 3. Substitute \((1, 45)\): \(45=b\cdot3\), so \(b=15\). Thus, \(f(x)=15\cdot3^x\). 4. Since \(3=e^{\ln(3)}\), \(f(x)=15e^{x\ln(3)}\). 5. Differentiate: \(f'(x)=15\ln(3)e^{x\ln(3)}\). Therefore, \(f'(0)=15\ln(3)\approx16.48\).

a) Exponential growth, because \(135>45\). b) \(a=3\) and \(b=15\) c) \(f(x)=15e^{x\ln(3)}\); \(f'(x)=15\ln(3)e^{x\ln(3)}\); \(f'(0)=15\ln(3)\approx16.48\)

- The first derivative gives the slope. - What is the maximum value of the cosine function? - Express the maximum slope as a function of \(k\).

1. Apply the chain rule: \(f_k'(x)=k\cos(kx)\). 2. Since the maximum value of cosine is \(1\), the maximum value of the derivative is \(k\). Therefore, the maximum slope is \(k\). 3. Replacing \(k\) by \(2k\) changes the maximum slope from \(k\) to \(2k\). 4. Thus, doubling \(k\) doubles the maximum slope.

The maximum slope is \(k\). When \(k\) is doubled, the maximum slope doubles to \(2k\).

- View each squared trigonometric expression as a composition. - Apply the power rule before differentiating the trigonometric inner function. - Remember to differentiate the linear expressions \(3x\) and \(2x\). - Look for a product that matches the double-angle identity.

1. Write \(f(x)=(\sin(3x))^2\). Applying the chain rule twice gives \(f'(x)=2\sin(3x)\cos(3x)\cdot 3\). Since \(2\sin(3x)\cos(3x)=\sin(6x)\), \(f'(x)=3\sin(6x)\). 2. Differentiate each term of \(g\): \(g'(x)=\frac{3}{2}[2\cos(x)(-\sin(x))]-\frac{3}{4}[-\sin(2x)\cdot 2]\) \(=-3\sin(x)\cos(x)+\frac{3}{2}\sin(2x)\). Using \(2\sin(x)\cos(x)=\sin(2x)\), \(g'(x)=-\frac{3}{2}\sin(2x)+\frac{3}{2}\sin(2x)=0\).

1) \(f'(x)=3\sin(6x)\) 2) \(g'(x)=0\)

- Use the product rule. - Apply the chain rule to the exponential factor. - Factor out the common exponential after differentiating. - Evaluate the first derivative to find the slope.

1. Let \(u(x)=4x-2\) and \(v(x)=e^{0.5x}\). Then \(u'(x)=4\) and \(v'(x)=0.5e^{0.5x}\). 2. Apply the product rule: \(g'(x)=4e^{0.5x}+(4x-2)(0.5)e^{0.5x}\). Factor and simplify: \(g'(x)=[4+0.5(4x-2)]e^{0.5x}=(2x+3)e^{0.5x}\). 3. Evaluate at \(x=0\): \(g'(0)=3e^0=3\).

a) \(g'(x)=(2x+3)e^{0.5x}\) b) The slope is \(3\).

- Set the two function values equal to find their intersection. - A substitution involving \(e^x\) can turn the intersection equation into a quadratic equation. - Compare both the function values and derivative values at the intersection. - Parallel lines have equal slopes.

1. Set the functions equal: \(e^x=2-e^{-x}\). Multiplying by \(e^x\) gives \(e^{2x}-2e^x+1=0\). Let \(u=e^x\). Then \((u-1)^2=0\), so \(u=1\), \(x=0\), and the intersection point is \(S(0, 1)\). 2. Differentiate: \(f'(x)=e^x\) and \(g'(x)=e^{-x}\). Thus \(f'(0)=g'(0)=1\), so the graphs share the tangent line \(y=x+1\) at \(S\). 3. To compare the graphs, \(f(x)-g(x)=e^x+e^{-x}-2\geq0\), with equality only at \(x=0\). Therefore, the graph of \(f\) stays above the graph of \(g\), and the graphs touch without crossing at \(S\). 4. At the same x-value, parallel tangent lines require \(f'(x)=g'(x)\). Solving \(e^x=e^{-x}\) gives \(e^{2x}=1\), so \(x=0\) is the only solution.

a) \(S(0, 1)\) b) \(f'(0)=g'(0)=1\). The graphs share the tangent line \(y=x+1\), and \(f(x)\geq g(x)\) with equality only at \(x=0\), so they do not cross. c) No. At the same x-value, the tangent lines are parallel only when \(x=0\).

- Decide which rule applies to each term. - Differentiate every exponent as part of the chain rule. - Use the product rule in part c. - Factor the common exponential factor when simplifying.

1. Differentiate each term in part a: \(f'(x)=8\cdot(-0.5)e^{-0.5x}+10x^4=-4e^{-0.5x}+10x^4\). 2. In part b, use the chain rule on both terms: \(g'(x)=e^{5x+2}+4\cos(4x)\). Here, \(\cos(-4x)=\cos(4x)\) because cosine is even. 3. In part c, use the product and chain rules: \(h'(x)=3e^{x^3}+(3x-4)(3x^2)e^{x^3}\) \(=(9x^3-12x^2+3)e^{x^3}\).

a) \(f'(x)=-4e^{-0.5x}+10x^4\) b) \(g'(x)=e^{5x+2}+4\cos(4x)\) c) \(h'(x)=(9x^3-12x^2+3)e^{x^3}\)

- Treat \(k\) as a constant. - A horizontal tangent has derivative value \(0\). - Substitute \(x=0\) after differentiating. - The exponential factor is never zero.

1. Apply the product and chain rules: \(f_k'(x)=ke^{-2x}-2(kx+2)e^{-2x}\). 2. Factor: \(f_k'(x)=(-2kx+k-4)e^{-2x}\). 3. A horizontal tangent at \(x=0\) requires \(f_k'(0)=0\). 4. Thus, \(k-4=0\), so \(k=4\).

a) \(f_k'(x)=(-2kx+k-4)e^{-2x}\) b) \(k=4\)

- Apply the quotient rule carefully. - Rewriting the denominator as a negative exponent turns the quotient into a product. - Use the chain rule for \(e^{ax}\). - A horizontal tangent occurs where the derivative is zero.

1. For the quotient-rule method, let \(u(x)=x+2\) and \(v(x)=e^{0.5x}\). Then \(u'(x)=1\) and \(v'(x)=0.5e^{0.5x}\). Thus, \(f'(x)=\frac{e^{0.5x}-0.5(x+2)e^{0.5x}}{e^x}\) \(=-0.5x e^{-0.5x}\). 2. For the product-rule method, \(f'(x)=e^{-0.5x}+(x+2)(-0.5e^{-0.5x})\) \(=-0.5x e^{-0.5x}\). 3. A horizontal tangent occurs where \(-0.5x e^{-0.5x}=0\). Since the exponential factor is never zero, \(x=0\).

a) \(f'(x)=-0.5x e^{-0.5x}\) b) The product-rule method gives the same expression. c) The horizontal tangent occurs at \(x=0\).

- Use the quotient rule for the overall function and the chain rule for the squared numerator. - Factor the derivative numerator instead of expanding it completely. - A horizontal tangent occurs where the derivative is zero. - A fraction is zero when its numerator is zero and its denominator is nonzero.

1. Let \(u(x)=(x^2-1)^2\) and \(v(x)=x^2+3\). By the chain rule, \(u'(x)=4x(x^2-1)\), and \(v'(x)=2x\). 2. Apply the quotient rule: \(f'(x)=\frac{4x(x^2-1)(x^2+3)-(x^2-1)^2(2x)}{(x^2+3)^2}\). 3. Factor \(2x(x^2-1)\) from the numerator: \(f'(x)=\frac{2x(x^2-1)[2(x^2+3)-(x^2-1)]}{(x^2+3)^2}\) \(=\frac{2x(x^2-1)(x^2+7)}{(x^2+3)^2}\). 4. A horizontal tangent occurs where the numerator is zero: \(2x(x^2-1)(x^2+7)=0\). Since \(x^2+7>0\), the real solutions are \(x=-1\), \(x=0\), and \(x=1\).

a) \(f'(x)=\frac{2x(x^2-1)(x^2+7)}{(x^2+3)^2}\) b) The horizontal tangents occur at \(x=-1\), \(x=0\), and \(x=1\).

- Check the domain of each original function before differentiating. - Identify the outer structure before choosing the product, quotient, or chain rule. - An endpoint containing a square root may need a difference-quotient check even when a chain-rule formula is valid only in the interior. - For a composite expression, account for the derivative of each inner function.

1. a) For \(x>0\), apply the product rule and chain rule: \(f'(x)=2xe^{\sqrt{x}}+x^2e^{\sqrt{x}}\frac{1}{2\sqrt{x}}=xe^{\sqrt{x}}\left(2+\frac{1}{2}\sqrt{x}\right)\). At the endpoint \(x=0\), the right-hand difference quotient is \(\lim_{h\to0^+}h e^{\sqrt{h}}=0\), so \(f'(0)=0\). Thus the derivative exists for \(x\ge0\). 2. b) For \(x>0\), apply the quotient rule and chain rule: \(f'(x)=\frac{\frac{1}{2\sqrt{x}}e^{x^2}-\sqrt{x}e^{x^2}(2x)}{e^{2x^2}}=\frac{1-4x^2}{2\sqrt{x}e^{x^2}}\). At \(x=0\), the right-hand difference quotient is unbounded, so the derivative does not exist there. Thus the derivative exists for \(x>0\). 3. c) Apply the chain rule: \(f'(x)=\frac{1}{2\sqrt{e^{3x}+1}}\cdot3e^{3x}=\frac{3e^{3x}}{2\sqrt{e^{3x}+1}}\). Since \(e^{3x}+1>0\) for every real \(x\), this derivative exists for all real \(x\).

a) \(f'(x)=xe^{\sqrt{x}}\left(2+\frac{1}{2}\sqrt{x}\right)\), for \(x\ge0\) b) \(f'(x)=\frac{1-4x^2}{2\sqrt{x}e^{x^2}}\), for \(x>0\) c) \(f'(x)=\frac{3e^{3x}}{2\sqrt{e^{3x}+1}}\), for \(x\in\mathbb{R}\)

- The derivative at a point gives the graph's slope there. - Rewrite \(\sqrt{e^x}\) as a single exponential expression before differentiating. - “Greater slope” means the larger derivative value. - Evaluate both derivatives at \(x=1\) and compare.

1. By the chain rule, \(f'(x)=e^{\sqrt{x}}\frac{1}{2\sqrt{x}}=\frac{e^{\sqrt{x}}}{2\sqrt{x}}\). 2. Since \(g(x)=e^{x/2}\), \(g'(x)=\frac{1}{2}e^{x/2}\). 3. At \(x=1\), \(f'(1)=\frac{e}{2}\) and \(g'(1)=\frac{\sqrt{e}}{2}\). Because \(e>\sqrt{e}\), \(f'(1)>g'(1)\).

\(f'(x)=\frac{e^{\sqrt{x}}}{2\sqrt{x}}\) and \(g'(x)=\frac{1}{2}e^{x/2}\). The graph of \(f\) has the greater slope at \(x=1\).

- Identify which function is the inner function in each composition. - Use the chain rule for a function of the form \(u(v(x))\). - Recall the derivative rules for logarithmic and power functions. - In part 2, the square applies to the entire logarithmic expression.

1. The first composition is \(h(x)=\ln(2x^2+5)\). By the chain rule, \(h'(x)=\frac{1}{2x^2+5}(4x)=\frac{4x}{2x^2+5}\). Since \(2x^2+5>0\) for every real \(x\), the domain of \(h\) is all real numbers. 2. The reverse composition is \(k(x)=2(\ln(x))^2+5\), which is defined for \(x>0\). Applying the chain rule gives \(k'(x)=4\ln(x)\left(\frac{1}{x}\right)=\frac{4\ln(x)}{x}\).

1) \(h(x)=\ln(2x^2+5)\), \(h'(x)=\frac{4x}{2x^2+5}\), and the domain of \(h\) is \((-\infty, \infty)\). 2) \(k(x)=2(\ln(x))^2+5\), \(k'(x)=\frac{4\ln(x)}{x}\), and the domain of \(k\) is \((0, \infty)\).

- Rewrite the expression as an outer power applied to an inner polynomial. - A horizontal tangent occurs where the derivative is zero. - Apply the power rule and then multiply by the derivative of the inner function. - A fraction equals zero when its numerator is zero and its denominator is nonzero.

1. One possible decomposition is \(u(t)=t^{-2}\) and \(v(x)=x^3-4\). 2. Since \(u'(t)=-2t^{-3}\) and \(v'(x)=3x^2\), the chain rule gives \(p'(x)=-2(x^3-4)^{-3}(3x^2)\) \(=-\frac{6x^2}{(x^3-4)^3}\). 3. A horizontal tangent occurs where \(p'(x)=0\). A rational expression is zero when its numerator is zero and its denominator is nonzero. Thus, \(-6x^2=0\), so \(x=0\). At \(x=0\), the denominator is nonzero, so this value is in the domain.

1) One possible choice is \(u(t)=t^{-2}\) and \(v(x)=x^3-4\). 2) \(p'(x)=-\frac{6x^2}{(x^3-4)^3}\) 3) \(x=0\)

- Use the quotient rule for the outer structure. - The radical in the numerator requires the chain rule. - Combine the numerator over a common radical denominator. - Distribute carefully when simplifying the numerator.

1. Let \(u(x)=\sqrt{x^2+1}\) and \(v(x)=2x-3\). Then \(u'(x)=\frac{x}{\sqrt{x^2+1}}\) and \(v'(x)=2\). 2. Apply the quotient rule: \(f'(x)=\frac{\frac{x}{\sqrt{x^2+1}}(2x-3)-2\sqrt{x^2+1}}{(2x-3)^2}\). 3. Combine the numerator over \(\sqrt{x^2+1}\): \(f'(x)=\frac{x(2x-3)-2(x^2+1)}{(2x-3)^2\sqrt{x^2+1}}=\frac{-3x-2}{(2x-3)^2\sqrt{x^2+1}}\).

\(f'(x)=\frac{-3x-2}{(2x-3)^2\sqrt{x^2+1}}\), where \(x\ne\frac{3}{2}\).

- Rewrite the cube root as a rational power. - Identify the outer, middle, and inner functions. - Remember the derivative of \(\sin(2x)\), including the derivative of its argument. - A negative rational exponent can be rewritten as a radical in the denominator.

1. Rewrite the function as \(f(x)=(\sin(2x)+2)^{\frac{1}{3}}\). 2. Apply the chain rule to the outer power and then to \(\sin(2x)\): \(f'(x)=\frac{1}{3}(\sin(2x)+2)^{-\frac{2}{3}}\cdot2\cos(2x)\). 3. Rewrite with a radical: \(f'(x)=\frac{2\cos(2x)}{3\sqrt[3]{(\sin(2x)+2)^2}}\).

\(f'(x)=\frac{2\cos(2x)}{3\sqrt[3]{(\sin(2x)+2)^2}}\)

- Use the product rule for the outer structure. - Differentiate the radical with the chain rule. - Combine the two terms over a common denominator.

1. Let \(u(x)=2x-3\) and \(v(x)=\sqrt{x^2+5}\). Then \(u'(x)=2\) and \(v'(x)=\frac{x}{\sqrt{x^2+5}}\). 2. Apply the product rule: \(f'(x)=2\sqrt{x^2+5}+(2x-3)\frac{x}{\sqrt{x^2+5}}\). 3. Combine over a common denominator: \(f'(x)=\frac{2(x^2+5)+x(2x-3)}{\sqrt{x^2+5}}=\frac{4x^2-3x+10}{\sqrt{x^2+5}}\).

\(f'(x)=\frac{4x^2-3x+10}{\sqrt{x^2+5}}\)

- Use the quotient rule for the outer structure. - Apply the chain rule to both \(e^{1-x}\) and \(\sqrt{x^2+1}\). - Factor the common exponential expression from the numerator. - Combine the radical factors in the denominator carefully.

1. Let \(u(x)=e^{1-x}\) and \(v(x)=\sqrt{x^2+1}\). Then \(u'(x)=-e^{1-x}\) and \(v'(x)=\frac{x}{\sqrt{x^2+1}}\). 2. Apply the quotient rule: \(g'(x)=\frac{-e^{1-x}\sqrt{x^2+1}-e^{1-x}\frac{x}{\sqrt{x^2+1}}}{x^2+1}\). 3. Combine and simplify: \(g'(x)=-\frac{(x^2+x+1)e^{1-x}}{(x^2+1)\sqrt{x^2+1}}\).

\(g'(x)=-\frac{(x^2+x+1)e^{1-x}}{(x^2+1)\sqrt{x^2+1}}\)

- Match each graph transformation to its parameter in the function formula. - A shift left changes the expression inside the radical to \(x+4\). - Differentiate the square-root function with the chain rule. - The derivative at \(x=5\) gives the tangent slope.

1. The transformations give \(a=3\), \(b=4\), and \(c=-2\), so \(f(x)=3\sqrt{x+4}-2\). 2. Differentiate: \(f'(x)=\frac{3}{2\sqrt{x+4}}\). 3. At \(x=5\), the slope is \(f'(5)=\frac{3}{2\sqrt{9}}=\frac{1}{2}\), and the point is \((5,7)\). 4. Using point-slope form, \(y-7=\frac{1}{2}(x-5)\), so \(y=\frac{1}{2}x+\frac{9}{2}\).

1. \(f(x)=3\sqrt{x+4}-2\) 2. \(y=\frac{1}{2}x+\frac{9}{2}\)

- Use radical properties to factor the expression before describing the transformations. - A factor outside the radical creates a vertical stretch. - The derivative gives the graph's slope. - Set the derivative equal to the target slope and solve.

1. Rewrite \(f(x)=\sqrt{4(x+3)}=2\sqrt{x+3}\). The graph is vertically stretched by a factor of \(2\) and shifted \(3\) units left. 2. Differentiate: \(f'(x)=\frac{1}{\sqrt{x+3}}\). 3. Set the derivative equal to \(1\): \(\frac{1}{\sqrt{x+3}}=1\). Thus, \(x+3=1\), so \(x_0=-2\).

1. \(f(x)=2\sqrt{x+3}\); vertical stretch by \(2\) and shift \(3\) units left 2. \(x_0=-2\)

- A square-root radicand must be nonnegative. - Determine where the downward-opening quadratic is nonnegative relative to its zeros. - Differentiating a square root places the radical in the denominator. - Division by zero is undefined.

1. The radicand must be nonnegative: \(12-x-x^2\ge0\). 2. Factor the corresponding quadratic: \(12-x-x^2=-(x+4)(x-3)\). It is nonnegative between its zeros, so \(D_f=[-4,3]\). 3. For interior points, \(f'(x)=\frac{-1-2x}{2\sqrt{12-x-x^2}}\). 4. This derivative is defined when the radicand is strictly positive, so \(D_{f'}=(-4,3)\). At \(x=-4\) and \(x=3\), the graph has vertical tangent behavior and no finite two-sided derivative.

a) \(D_f=[-4,3]\) b) \(D_{f'}=(-4,3)\)

- A square root in a denominator must be positive, not merely nonnegative. - Use the quotient rule and the chain rule. - Combine the numerator terms over a common denominator. - A horizontal tangent requires the first derivative to equal zero.

1. Because the square root is in the denominator, \(x-2\) must be strictly positive. Thus the domain is \((2,\infty)\). 2. Let \(u(x)=x+2\) and \(v(x)=\sqrt{x-2}\). Then \(u'(x)=1\) and \(v'(x)=\frac{1}{2\sqrt{x-2}}\). 3. The quotient rule gives \(f'(x)=\frac{\sqrt{x-2}-(x+2)\frac{1}{2\sqrt{x-2}}}{x-2}\). Combining terms yields \(f'(x)=\frac{x-6}{2\sqrt{(x-2)^3}}=\frac{0.5x-3}{\sqrt{(x-2)^3}}\). 4. A horizontal tangent requires \(f'(x)=0\). The denominator is positive on the domain, so \(0.5x-3=0\), giving \(x=6\). 5. Since \(f(6)=\frac{8}{2}=4\), the point is \((6,4)\).

a) \((2,\infty)\). b) \(f'(x)=\frac{0.5x-3}{\sqrt{(x-2)^3}}\). c) \((6,4)\).

- Use the quotient rule and apply the chain rule to the square root. - Combine the two terms in the numerator over a common denominator. - A horizontal tangent requires the first derivative to be zero. - Check whether the numerator of the derivative can ever be zero.

1. Let \(u(x)=\sqrt{x^2+12}\) and \(v(x)=x\). Then \(u'(x)=\frac{x}{\sqrt{x^2+12}}\) and \(v'(x)=1\). 2. The quotient rule gives \(g'(x)=\frac{\frac{x}{\sqrt{x^2+12}}\cdot x-\sqrt{x^2+12}}{x^2}\). 3. Combining the numerator over a common denominator gives \(\frac{x^2-(x^2+12)}{\sqrt{x^2+12}}=\frac{-12}{\sqrt{x^2+12}}\). Therefore, \(g'(x)=-\frac{12}{x^2\sqrt{x^2+12}}\). 4. A horizontal tangent would require \(g'(x)=0\). The numerator is the nonzero constant \(-12\), while the denominator is positive on the domain, so the derivative is never zero.

a) \(g'(x)=-\frac{12}{x^2\sqrt{x^2+12}}\). b) The derivative has constant numerator \(-12\neq0\), so it has no zeros and the graph has no horizontal tangents.

- Require each square-root radicand to be nonnegative. - Set the function values equal to find an intersection. - After squaring, check the resulting solution in the original equation. - The derivative at the intersection gives the tangent slope. - Use point-slope form for the tangent line.

1. The radicands must be nonnegative, so \(D_f=[-4,\infty)\) and \(D_g=[0,\infty)\). 2. Set the functions equal: \(\sqrt{x+4}=4-\sqrt{x}\). Squaring and simplifying gives \(8\sqrt{x}=12\), so \(\sqrt{x}=\frac{3}{2}\) and \(x=\frac{9}{4}\). Then \(y=\frac{5}{2}\), so \(S=\left(\frac{9}{4},\frac{5}{2}\right)\). 3. Since \(f'(x)=\frac{1}{2\sqrt{x+4}}\), the tangent slope is \(f'\left(\frac{9}{4}\right)=\frac{1}{5}\). 4. Using point-slope form, \(y-\frac{5}{2}=\frac{1}{5}\left(x-\frac{9}{4}\right)\), so \(y=\frac{1}{5}x+\frac{41}{20}\).

a) \(D_f=[-4,\infty)\); \(D_g=[0,\infty)\) b) \(S=\left(\frac{9}{4},\frac{5}{2}\right)\) c) \(y=\frac{1}{5}x+\frac{41}{20}\)

- Set \(y=0\) for x-intercepts and \(x=0\) for y-intercepts. - Set the two function expressions equal to find shared points. - Apply the chain rule to each square-root function. - Two nonvertical lines are perpendicular when the product of their slopes is \(-1\).

1. a) For \(f\), \(f(0)=0\), so the only axis intercept is \((0,0)\). For \(g\), \(g(0)=\sqrt{6}\), giving \((0,\sqrt{6})\), and \(g(6)=0\), giving \((6,0)\). 2. b) Set the functions equal: \(\sqrt{2x}=\sqrt{6-x}\). Squaring gives \(2x=6-x\), so \(x=2\) and \(y=2\). The resulting linear equation has only one solution, and it checks in the original equation. 3. c) The derivatives are \(f'(x)=\frac{1}{\sqrt{2x}}\) and \(g'(x)=-\frac{1}{2\sqrt{6-x}}\). 4. At \(x=2\), the slopes are \(m_f=\frac{1}{2}\) and \(m_g=-\frac{1}{4}\). Their product is \(-\frac{1}{8}\ne-1\), so the tangent lines are not perpendicular.

a) For \(f\): \((0,0)\); for \(g\): \((0,\sqrt{6})\) and \((6,0)\) b) The only shared point is \((2,2)\). c) No. The tangent slopes have product \(-\frac{1}{8}\), not \(-1\).

- A square-root radicand must be nonnegative. - A tangent line requires both the point on the graph and the derivative there. - Use the chain rule to differentiate the square root. - A point on the y-axis has \(x=0\).

1. a) Require \(6-3x\ge0\), which gives \(x\le2\). Thus, \(D_f=(-\infty,2]\). 2. b) The point of tangency is \((-1,f(-1))=(-1,5)\). 3. Differentiate: \(f'(x)=-\frac{3}{2\sqrt{6-3x}}\), defined for \(x<2\). Therefore, \(f'(-1)=-\frac{1}{2}\). 4. Using point-slope form, \(y-5=-\frac{1}{2}(x+1)\), so \(t\) is \(y=-\frac{1}{2}x+\frac{9}{2}\). 5. c) Set \(x=0\). The y-intercept is \(\left(0,\frac{9}{2}\right)\).

a) \(D_f=(-\infty,2]\) b) \(t:y=-\frac{1}{2}x+\frac{9}{2}\) c) \(\left(0,\frac{9}{2}\right)\)

- Require the expression under the square root to be nonnegative. - Set the derivative equal to the specified tangent slope. - To solve a reciprocal square-root equation, isolate the radical before squaring. - A line is determined by a slope and one point.

1. a) Require \(2x+1\ge0\), so \(D_k=\left[-\frac{1}{2},\infty\right)\). 2. Differentiate: \(k'(x)=-\frac{1}{\sqrt{2x+1}}\). 3. Set \(k'(x_0)=-\frac{1}{5}\): \(\sqrt{2x_0+1}=5\), so \(2x_0+1=25\) and \(x_0=12\). 4. The point of tangency is \((12,k(12))=(12,-1)\). 5. Using point-slope form, \(y+1=-\frac{1}{5}(x-12)\), so \(y=-\frac{1}{5}x+\frac{7}{5}\).

a) \(D_k=\left[-\frac{1}{2},\infty\right)\) b) \(x_0=12\) c) \(y=-\frac{1}{5}x+\frac{7}{5}\)

- A square-root radicand must be nonnegative. - Apply the chain rule to the square-root function. - A tangent line needs the function value and derivative value at the given input. - Use point-slope form after finding the slope.

1. a) Require \(12-3x^2\ge0\), or \(x^2\le4\). Therefore, \(D_f=[-2,2]\). 2. b) By the chain rule, \(f'(x)=\frac{-6x}{2\sqrt{12-3x^2}}=-\frac{3x}{\sqrt{12-3x^2}}\). 3. c) At \(x=1\), \(f(1)=3\) and \(f'(1)=-1\). 4. Using point-slope form, \(y-3=-(x-1)\), so \(y=-x+4\).

a) \(D_f=[-2,2]\) b) \(f'(x)=-\frac{3x}{\sqrt{12-3x^2}}\) c) \(y=-x+4\)

- Identify which operations occur inside and outside the radical. - Find an x-intercept by setting the function equal to \(0\). - Isolate the radical before squaring. - Differentiate the square-root function to obtain its slope function. - Set the derivative equal to the specified slope.

1. a) Require \(x+1\ge0\), so \(D_h=[-1,\infty)\). Starting with \(w(x)=\sqrt{x}\), shift \(1\) unit left, vertically stretch by \(2\), reflect across the x-axis, and shift \(3\) units up. 2. b) Set \(h(x)=0\): \(3-2\sqrt{x+1}=0\), so \(\sqrt{x+1}=\frac{3}{2}\). Then \(x+1=\frac{9}{4}\), giving \(x=\frac{5}{4}\). The intercept is \(\left(\frac{5}{4},0\right)\). 3. c) Differentiate: \(h'(x)=-\frac{1}{\sqrt{x+1}}\). Set \(h'(x_0)=-\frac{1}{2}\), so \(\sqrt{x_0+1}=2\), and \(x_0=3\).

a) \(D_h=[-1,\infty)\); shift left \(1\), vertically stretch by \(2\), reflect across the x-axis, and shift up \(3\) b) \(\left(\frac{5}{4},0\right)\) c) \(x_0=3\)

- Match each parameter to a horizontal shift, vertical scale, or vertical shift. - Rewrite the square root as a rational power before differentiating if helpful. - Parallel lines have equal slopes. - Substitute the x-coordinate into \(f\) to find the point's y-coordinate.

1. a) Shift the square-root graph \(4\) units left, vertically shrink it by a factor of \(\frac{1}{2}\), and shift it \(3\) units down. 2. b) By the chain rule, \(f'(x)=\frac{1}{4\sqrt{x+4}}\). 3. c) Parallel lines have equal slopes, so set \(f'(x)=\frac{1}{8}\). Then \(\frac{1}{4\sqrt{x+4}}=\frac{1}{8}\), so \(\sqrt{x+4}=2\), giving \(x=0\). 4. Since \(f(0)=\frac{1}{2}\sqrt{4}-3=-2\), the point is \(P=(0,-2)\).

a) Shift left \(4\), vertically shrink by \(\frac{1}{2}\), and shift down \(3\) b) \(f'(x)=\frac{1}{4\sqrt{x+4}}\) c) \(P=(0,-2)\)

- Test \(f(-x)=f(x)\) for y-axis symmetry and \(f(-x)=-f(x)\) for origin symmetry. - Replace \(x\) with \(-x\) and compare expressions. - Use the quotient rule or rewrite the denominator with a negative exponent. - Squaring removes the sign of \(x\).

1. Evaluate \(f(-x)=\frac{-x}{\sqrt{x^2+4}}=-f(x)\). Therefore, \(f\) is odd, and its graph is symmetric about the origin. 2. Differentiate with the quotient rule: \(f'(x)=\frac{\sqrt{x^2+4}-x\left(\frac{x}{\sqrt{x^2+4}}\right)}{x^2+4}=\frac{4}{(x^2+4)\sqrt{x^2+4}}\). 3. Since \(f'(-x)=\frac{4}{(x^2+4)\sqrt{x^2+4}}=f'(x)\), the derivative is even, and its graph is symmetric about the y-axis.

The graph of \(f\) is symmetric about the origin. \(f'(x)=\frac{4}{(x^2+4)\sqrt{x^2+4}}\), and the graph of \(f'\) is symmetric about the y-axis.

- Even powers do not change when \(x\) is replaced by \(-x\). - Use the chain rule for the square-root function. - A smooth graph symmetric about the y-axis has a horizontal tangent where it crosses that axis. - Determine the parity of the derivative of an even function.

1. a) \(g(-x)=\sqrt{(-x)^4+2(-x)^2+5}=\sqrt{x^4+2x^2+5}=g(x)\). Thus, \(g\) is even, so its graph is symmetric about the y-axis. 2. b) By the chain rule, \(g'(x)=\frac{4x^3+4x}{2\sqrt{x^4+2x^2+5}}=\frac{2x^3+2x}{\sqrt{x^4+2x^2+5}}\). 3. The derivative of an even differentiable function is odd. Therefore, \(g'(-x)=-g'(x)\), and at \(x=0\), \(g'(0)=-g'(0)\), which implies \(g'(0)=0\).

a) Since \(g(-x)=g(x)\), the graph is symmetric about the y-axis. b) \(g'(x)=\frac{2x^3+2x}{\sqrt{x^4+2x^2+5}}\), and \(g'(0)=0\) because the derivative of an even differentiable function is odd.

- Use the quotient rule. - Differentiate the square root with the chain rule. - Combine the terms in the numerator over a common denominator. - Use exponent rules to combine \((x+1)\sqrt{x+1}\).

1. Let \(u(x)=x+3\) and \(v(x)=\sqrt{x+1}\). Then \(u'(x)=1\) and \(v'(x)=\frac{1}{2\sqrt{x+1}}\). 2. Apply the quotient rule: \(f'(x)=\frac{\sqrt{x+1}-(x+3)\frac{1}{2\sqrt{x+1}}}{x+1}\). 3. Combine the numerator: \(f'(x)=\frac{2(x+1)-(x+3)}{2\sqrt{x+1}(x+1)}=\frac{x-1}{2\sqrt{x+1}(x+1)}\). 4. Since \((x+1)\sqrt{x+1}=\sqrt{(x+1)^3}\), \(f'(x)=\frac{x-1}{2\sqrt{(x+1)^3}}\).

\(f'(x)=\frac{x-1}{2\sqrt{(x+1)^3}}\)

- Apply the chain rule when differentiating the radical in the denominator. - Use the quotient rule for the outer structure. - Combine the numerator over a common radical denominator. - Look for terms that cancel after expansion.

1. Let \(u(x)=4x\) and \(v(x)=\sqrt{x^2+3}\). Then \(u'(x)=4\) and \(v'(x)=\frac{x}{\sqrt{x^2+3}}\). 2. Apply the quotient rule: \(g'(x)=\frac{4\sqrt{x^2+3}-4x\frac{x}{\sqrt{x^2+3}}}{x^2+3}\). 3. Combine the numerator: \(g'(x)=\frac{4(x^2+3)-4x^2}{(x^2+3)\sqrt{x^2+3}}=\frac{12}{(x^2+3)\sqrt{x^2+3}}\).

\(g'(x)=\frac{12}{(x^2+3)\sqrt{x^2+3}}\)

- A square-root radicand must be nonnegative. - Set a function equal to \(0\) to find an x-intercept. - Set the two functions equal to find a shared point. - Use each derivative to find the corresponding tangent slope. - Write each tangent line in point-slope form.

1. a) Require each radicand to be nonnegative. Thus, \(D_f=(-\infty,12]\) and \(D_g=[-4,\infty)\). The x-intercepts are \((12,0)\) and \((-4,0)\), respectively. 2. b) Set the functions equal: \(12-x=x+4\), so \(x=4\). Then \(y=\sqrt{8}=2\sqrt{2}\), giving \(S=(4,2\sqrt{2})\). 3. c) The derivatives are \(f'(x)=-\frac{1}{2\sqrt{12-x}}\) and \(g'(x)=\frac{1}{2\sqrt{x+4}}\). 4. At \(x=4\), the slopes are \(-\frac{\sqrt{2}}{8}\) and \(\frac{\sqrt{2}}{8}\). 5. Using point-slope form, the tangent lines are \(t_f:y=-\frac{\sqrt{2}}{8}x+\frac{5\sqrt{2}}{2}\) and \(t_g:y=\frac{\sqrt{2}}{8}x+\frac{3\sqrt{2}}{2}\).

a) \(D_f=(-\infty,12]\), with x-intercept \((12,0)\); \(D_g=[-4,\infty)\), with x-intercept \((-4,0)\) b) \(S=(4,2\sqrt{2})\) c) \(t_f:y=-\frac{\sqrt{2}}{8}x+\frac{5\sqrt{2}}{2}\); \(t_g:y=\frac{\sqrt{2}}{8}x+\frac{3\sqrt{2}}{2}\)

- Require the square-root radicand to be nonnegative. - Find the graph point at the given x-coordinate. - Apply the chain rule to the radical term. - A line is determined by one point and its slope.

1. a) Require \(7-x\ge0\), so \(D=(-\infty,7]\). 2. At \(x=3\), \(f(3)=2\sqrt{4}+3=7\). 3. By the chain rule, \(f'(x)=1-\frac{1}{\sqrt{7-x}}\), so \(f'(3)=\frac{1}{2}\). 4. Using point-slope form, \(y-7=\frac{1}{2}(x-3)\), so \(y=\frac{1}{2}x+\frac{11}{2}\).

a) \(D=(-\infty,7]\) b) \(y=\frac{1}{2}x+\frac{11}{2}\)

- Require the expression under the radical to be nonnegative. - A tangent slope is a value of the derivative. - Set the derivative equal to the specified slope and solve. - Substitute the resulting x-coordinate into the original function.

1. a) Require \(2x+2\ge0\), so the domain is \([-1,\infty)\). 2. Differentiate: \(g'(x)=\frac{3}{\sqrt{2x+2}}-1\). 3. Set the derivative equal to \(\frac{1}{2}\): \(\frac{3}{\sqrt{2x+2}}-1=\frac{1}{2}\). Then \(\sqrt{2x+2}=2\), so \(x=1\). 4. Evaluate \(g(1)=3\sqrt{4}-1=5\). Thus, \(P=(1,5)\).

a) \([-1,\infty)\) b) \(P=(1,5)\)

- Set \(x=0\) for the y-intercept and \(y=0\) for the x-intercept. - The derivative gives the tangent slope. - Relate the slope to the tangent of the inclination angle. - Read transformations from the operations inside and outside the square root.

1. a) The y-intercept is \((0,g(0))=(0,2)\). For the x-intercept, solve \(4-2\sqrt{x+1}=0\), giving \(x=3\). Thus, the x-intercept is \((3,0)\). 2. b) Differentiate: \(g'(x)=-\frac{1}{\sqrt{x+1}}\). At \(x=0\), the tangent slope is \(-1\). Since \(\tan(135^\circ)=-1\), the inclination angle is \(135^\circ\). 3. c) Starting with \(w(x)=\sqrt{x}\), shift \(1\) unit left, vertically stretch by \(2\), reflect across the x-axis, and shift \(4\) units up.

a) y-intercept \((0,2)\); x-intercept \((3,0)\) b) \(g'(0)=-1=\tan(135^\circ)\), so the inclination angle is \(135^\circ\). c) Shift left \(1\), vertically stretch by \(2\), reflect across the x-axis, and shift up \(4\).

- Rewrite the square root as a power before applying the chain rule. - The derivative's denominator determines where the derivative is defined. - For the one-sided limit, track the sign and size of the denominator as the endpoint is approached.

1. Write \(f(x)=(10-2x)^{1/2}-2\). By the chain rule, \(f'(x)=\frac{1}{2}(10-2x)^{-1/2}(-2)=-\frac{1}{\sqrt{10-2x}}\). 2. The derivative is defined when \(10-2x>0\), so the domain of \(f'\) is \(( -\infty,5)\). 3. As \(x\to5^-\), the positive denominator \(\sqrt{10-2x}\) approaches \(0\), so \(f'(x)\to-\infty\). 4. The tangent direction becomes vertical as the graph approaches the endpoint \((5,-2)\), so the graph has a vertical tangent there.

a) \(f'(x)=-\frac{1}{\sqrt{10-2x}}\), with domain \(( -\infty,5)\) b) \(\lim_{x\to5^-}f'(x)=-\infty\) c) The graph has a vertical tangent at the endpoint \((5,-2)\).

- The normal slope is the negative reciprocal of the tangent slope. - Set the normal line equal to \(0\) to find its x-intercept. - Compare the resulting x-coordinate with \(x_0\). - A point on the x-axis is as far from the origin as the absolute value of its x-coordinate.

1. Differentiate: \(g'(x)=\frac{1}{\sqrt{2x}}\). 2. The normal slope at \(x_0\) is \(m_n=-\sqrt{2x_0}\). 3. The normal equation is \(y=-\sqrt{2x_0}(x-x_0)+\sqrt{2x_0}\). 4. a) Set \(y=0\): \(0=-\sqrt{2x_0}(x_n-x_0)+\sqrt{2x_0}\). Since \(x_0>0\), divide by \(\sqrt{2x_0}\) to obtain \(x_n=x_0+1\). 5. b) For \(x_0=4\), \(x_n=5\), so \(S=(5,0)\). The distance \(OS\) is \(5\) units.

a) \(x_n=x_0+1\) b) \(S=(5,0)\), and \(OS=5\) units

- A y-intercept has x-coordinate \(0\). - The tangent slope is the derivative value at the point. - Relate a line's slope to the tangent of its inclination angle. - The normal slope is the negative reciprocal of the tangent slope.

1. The y-intercept is \(P=(0,h(0))=(0,6)\). 2. By the chain rule, \(h'(x)=-\frac{3}{2\sqrt{4-x}}\), so the tangent slope at \(x=0\) is \(m_t=-\frac{3}{4}\). 3. The inclination angle \(\alpha\) satisfies \(\tan(\alpha)=-\frac{3}{4}\) with \(0^\circ<\alpha<180^\circ\). Thus, \(\alpha\approx143.13^\circ\). 4. The normal slope is \(m_n=-\frac{1}{m_t}=\frac{4}{3}\). Through \((0,6)\), the normal line is \(y=\frac{4}{3}x+6\).

Normal line: \(y=\frac{4}{3}x+6\) Tangent inclination angle: \(\alpha\approx143.13^\circ\)

- The radicand must be nonnegative. - Use the chain rule for the square root. - The derivative value gives the slope. - Isolate and square a radical equation only after checking signs.

1. Require \(2ax-a^2\geq0\). Since \(a>0\), this is equivalent to \(x\geq\frac{a}{2}\). Thus, the domain is \(\left[\frac{a}{2},\infty\right)\). 2. Apply the chain rule: \(f_a'(x)=\frac{a}{\sqrt{2ax-a^2}}\). 3. Set \(f_a'(2)=\frac{1}{2}\): \(\frac{a}{\sqrt{4a-a^2}}=\frac{1}{2}\). 4. Squaring gives \(a^2=\frac{1}{4}(4a-a^2)\), so \(5a^2-4a=0\). 5. Since \(a>0\), \(a=\frac{4}{5}\). The value \(x=2\) lies in the corresponding domain, so the solution is valid.

a) Domain: \(\left[\frac{a}{2},\infty\right)\) b) \(f_a'(x)=\frac{a}{\sqrt{2ax-a^2}}\) c) \(a=\frac{4}{5}\)

- Require the square-root radicand to be nonnegative. - Use the zeros of the quadratic to make a sign chart. - Apply the chain rule to the composite function. - Simplify the factor of \(2\) after differentiating.

1. Require \(x^2-6x+5\ge0\). Since \(x^2-6x+5=(x-1)(x-5)\), the expression is nonnegative for \(x\le1\) or \(x\ge5\). 2. Therefore, \(D_f=(-\infty,1]\cup[5,\infty)\). 3. By the chain rule, \(f'(x)=\frac{2x-6}{2\sqrt{x^2-6x+5}}=\frac{x-3}{\sqrt{x^2-6x+5}}\), for \(x<1\) or \(x>5\).

\(D_f=(-\infty,1]\cup[5,\infty)\); \(f'(x)=\frac{x-3}{\sqrt{x^2-6x+5}}\) on the interior of \(D_f\)

- Check both the square-root condition and the denominator restriction. - Apply the quotient rule. - Differentiate the radical in the numerator with the chain rule. - Combine the resulting numerator over a common radical denominator.

1. The radicand condition gives \(x\ge-\frac{1}{4}\), and the denominator requires \(x\ne0\). Thus, \(D_f=\left[-\frac{1}{4},0\right)\cup(0,\infty)\). 2. Let \(u(x)=\sqrt{4x+1}\) and \(v(x)=x\). Then \(u'(x)=\frac{2}{\sqrt{4x+1}}\) and \(v'(x)=1\). 3. By the quotient rule, \(f'(x)=\frac{\frac{2x}{\sqrt{4x+1}}-\sqrt{4x+1}}{x^2}\). 4. Combine the numerator: \(f'(x)=\frac{2x-(4x+1)}{x^2\sqrt{4x+1}}=\frac{-2x-1}{x^2\sqrt{4x+1}}\).

\(D_f=\left[-\frac{1}{4},0\right)\cup(0,\infty)\); \(f'(x)=\frac{-2x-1}{x^2\sqrt{4x+1}}\) on the interior of \(D_f\)

- Read horizontal changes from inside the radical and vertical changes from outside it. - Apply the chain rule to differentiate. - Substitute the given input into the derivative to find the slope. - Use the endpoint value and long-run behavior to determine the range.

1. Shift the square-root graph \(4\) units left, vertically stretch it by \(2\), reflect it across the x-axis, and shift it \(3\) units up. 2. By the chain rule, \(f'(x)=-\frac{1}{\sqrt{x+4}}\), for \(x>-4\). Thus, \(f'(0)=-\frac{1}{2}\). 3. At the left endpoint, \(f(-4)=3\), which is the greatest value. As \(x\to\infty\), \(f(x)\to-\infty\). Therefore, the range is \((-\infty,3]\).

1. Shift left \(4\), vertically stretch by \(2\), reflect across the x-axis, and shift up \(3\). 2. \(f'(x)=-\frac{1}{\sqrt{x+4}}\); \(f'(0)=-\frac{1}{2}\) 3. \((-\infty,3]\)

- Require the square-root radicand to be nonnegative. - Use the product rule for a linear factor times a radical factor. - Apply the chain rule to the radical. - Combine the two derivative terms over a common denominator.

1. Require \(6-x\ge0\), so \(D=(-\infty,6]\). 2. Let \(u(x)=x+4\) and \(v(x)=\sqrt{6-x}\). Then \(u'(x)=1\) and \(v'(x)=-\frac{1}{2\sqrt{6-x}}\). 3. By the product rule, \(f'(x)=\sqrt{6-x}-\frac{x+4}{2\sqrt{6-x}}\). 4. Combine over a common denominator: \(f'(x)=\frac{2(6-x)-(x+4)}{2\sqrt{6-x}}=\frac{8-3x}{2\sqrt{6-x}}\), for \(x<6\).

\(D=(-\infty,6]\); \(f'(x)=\frac{8-3x}{2\sqrt{6-x}}\) for \(x<6\)

- Parallel lines have equal slopes. - Set the derivative equal to the target slope. - Apply the chain rule to the square-root function. - Check solutions after squaring because signs may have been lost. - Use point-slope form once the point of tangency is known.

1. Differentiate: \(f'(x)=-\frac{x}{\sqrt{100-x^2}}\). 2. A parallel tangent must have slope \(\frac{3}{4}\), so solve \(-\frac{x}{\sqrt{100-x^2}}=\frac{3}{4}\). 3. Squaring gives \(16x^2=9(100-x^2)\), so \(x^2=36\). Checking the original equation shows that the positive slope requires \(x=-6\). 4. The point of tangency is \((-6,f(-6))=(-6,8)\). 5. Using point-slope form, \(y-8=\frac{3}{4}(x+6)\), so \(y=\frac{3}{4}x+\frac{25}{2}\).

\(y=\frac{3}{4}x+\frac{25}{2}\)

- Perpendicular nonvertical lines have slopes whose product is \(-1\). - Set the derivative equal to the required tangent slope. - Check the sign after squaring. - Use the point and slope to write the line equation.

1. Differentiate: \(g'(x)=\frac{x}{\sqrt{x^2+12}}\). 2. A line perpendicular to a line of slope \(-2\) has slope \(\frac{1}{2}\). 3. Solve \(\frac{x}{\sqrt{x^2+12}}=\frac{1}{2}\). Squaring gives \(4x^2=x^2+12\), so \(x^2=4\). Checking the original equation gives \(x=2\). 4. The point of tangency is \((2,g(2))=(2,4)\). 5. Using point-slope form, \(y-4=\frac{1}{2}(x-2)\), so \(y=\frac{1}{2}x+3\).

\(y=\frac{1}{2}x+3\)

- A square-root radicand must be nonnegative. - Apply the chain rule: outer derivative times inner derivative. - The tangent slope at a point is the derivative value there.

1. a) Require \(25-x^2\ge0\), so \(D=[-5,5]\). 2. b) Apply the chain rule: \(f'(x)=2\cdot\frac{1}{2\sqrt{25-x^2}}(-2x)=-\frac{2x}{\sqrt{25-x^2}}\), for \(-5<x<5\). 3. c) \(f'(3)=-\frac{6}{\sqrt{16}}=-\frac{3}{2}\).

a) \(D=[-5,5]\) b) \(f'(x)=-\frac{2x}{\sqrt{25-x^2}}\) for \(-5<x<5\) c) \(-\frac{3}{2}\)

- Apply the chain rule and multiply by the derivative of the radicand. - A tangent line requires the graph point and derivative value. - Use point-slope form to write the line.

1. a) By the chain rule, \(g'(x)=\frac{4x}{2\sqrt{2x^2+17}}=\frac{2x}{\sqrt{2x^2+17}}\). 2. b) At \(x=2\), \(g(2)=5\) and \(g'(2)=\frac{4}{5}\). 3. Using point-slope form, \(y-5=\frac{4}{5}(x-2)\), so \(y=\frac{4}{5}x+\frac{17}{5}\).

a) \(g'(x)=\frac{2x}{\sqrt{2x^2+17}}\) b) \(y=\frac{4}{5}x+\frac{17}{5}\)

- Require the logarithm’s input to be positive. - Differentiate the outer logarithm and multiply by the inner derivative. - Substitute \(x=2\) into the derivative.

1. The logarithm requires \(12-4x>0\). Solving gives \(x<3\), so \(D_g=(-\infty, 3)\). 2. Apply the chain rule: \(g'(x)=\frac{-4}{12-4x}=\frac{1}{x-3}\). 3. Evaluate the derivative at \(x=2\): \(g'(2)=\frac{1}{2-3}=-1\).

a) \(D_g=(-\infty, 3)\) b) \(g'(x)=\frac{1}{x-3}\) c) The slope is \(-1\).

- Determine when the logarithm’s argument is positive. - Apply the chain rule as outer derivative times inner derivative. - Compare the domains of \(\ln(x^4)\) and \(\ln x\). - Consider how absolute value produces a valid identity for negative inputs.

1. The logarithm requires \(x^4>0\), which holds exactly when \(x\ne0\). Therefore, \(D_f=\mathbb{R}\setminus\{0\}\). 2. Apply the chain rule: \(f'(x)=\frac{1}{x^4}(4x^3)=\frac{4}{x}\), for \(x\ne0\). 3. The derivative formula is correct on the entire domain, but the stated identity is not. The expression \(4\ln x\) is defined only for \(x>0\), while \(\ln(x^4)\) is also defined for \(x<0\). A domain-valid identity is \(\ln(x^4)=4\ln|x|\) for \(x\ne0\).

a) \(D_f=\mathbb{R}\setminus\{0\}\) b) \(f'(x)=\frac{4}{x}\), for \(x\ne0\) c) The derivative result is correct, but the identity \(\ln(x^4)=4\ln x\) is valid only for \(x>0\). On the full domain, use \(\ln(x^4)=4\ln|x|\).

- Find where the squared expression equals zero. - Apply the chain rule carefully. - A tangent line requires a point and a slope. - Evaluate both the function and derivative at \(x=1\).

1. The logarithm’s argument is zero only when \(2x-4=0\), so \(x=2\). It is positive for every other real input. Therefore, \(D_k=\mathbb{R}\setminus\{2\}\). 2. Differentiate using the chain rule: \(k'(x)=\frac{4(2x-4)}{(2x-4)^2}=\frac{2}{x-2}\). At \(x=1\), \(k(1)=\ln4\) and \(k'(1)=-2\). Thus, the tangent line is \(y=-2(x-1)+\ln4=-2x+2+\ln4\).

a) \(D_k=\mathbb{R}\setminus\{2\}\) b) \(y=-2x+2+\ln4\)

- Distinguish between composing functions and combining their outputs arithmetically. - Identify the inner and outer functions in each composition. - Decide which rules are needed when functions are multiplied or added. - Find the derivatives of the three basic functions first.

1. The functions can be written as \(f_1(x)=(u\circ w)(x)v(x)\), \(f_2(x)=(v\circ w)(x)+u(x)\), and \(f_3(x)=\frac{w(x)}{u(x)}\). 2. The basic derivatives are \(u'(x)=e^x\), \(v'(x)=-\sin(x)\), and \(w'(x)=2x\). 3. For \(f_1\), use the chain rule and product rule: \(f_1'(x)=e^{x^2+3}(2x)\cos(x)+e^{x^2+3}(-\sin(x))\) \(=e^{x^2+3}(2x\cos(x)-\sin(x))\). 4. For \(f_2\), use the chain rule and sum rule: \(f_2'(x)=-\sin(x^2+3)(2x)+e^x\) \(=e^x-2x\sin(x^2+3)\).

a) \(f_1(x)=(u\circ w)(x)v(x)\) b) \(f_2(x)=(v\circ w)(x)+u(x)\) c) \(f_3(x)=\frac{w(x)}{u(x)}\) \(u'(x)=e^x\), \(v'(x)=-\sin(x)\), and \(w'(x)=2x\) \(f_1'(x)=e^{x^2+3}(2x\cos(x)-\sin(x))\) \(f_2'(x)=e^x-2x\sin(x^2+3)\)

- Identify which function is used as the input to another function. - Use the chain rule: derivative of the outer function times derivative of the inner function. - Recall the derivatives of the natural logarithm and square root. - The third function has a quotient structure.

1. The functions can be written as \(g_1(x)=(a\circ c)(x)\), \(g_2(x)=(b\circ a)(x)\), and \(g_3(x)=\frac{b(x)}{c(x)}\). 2. The basic derivatives are \(a'(x)=\frac{1}{x}\), \(b'(x)=\frac{1}{2\sqrt{x}}\), and \(c'(x)=-2x\). 3. For \(g_1\), the chain rule gives \(g_1'(x)=\frac{1}{5-x^2}(-2x)=\frac{2x}{x^2-5}\). 4. For \(g_2\), the chain rule gives \(g_2'(x)=\frac{1}{2\sqrt{\ln(x)}}\left(\frac{1}{x}\right)=\frac{1}{2x\sqrt{\ln(x)}}\). The domains are \((-\sqrt{5}, \sqrt{5})\) for \(g_1\), \([1, \infty)\) for \(g_2\), and \([0, \sqrt{5})\cup(\sqrt{5}, \infty)\) for \(g_3\). The derivative \(g_2'\) is defined for \(x>1\).

a) \(g_1(x)=(a\circ c)(x)\) b) \(g_2(x)=(b\circ a)(x)\) c) \(g_3(x)=\frac{b(x)}{c(x)}\) \(a'(x)=\frac{1}{x}\), \(b'(x)=\frac{1}{2\sqrt{x}}\), and \(c'(x)=-2x\) \(g_1'(x)=\frac{2x}{x^2-5}\) for \(-\sqrt{5}<x<\sqrt{5}\) \(g_2'(x)=\frac{1}{2x\sqrt{\ln(x)}}\) for \(x>1\)

- The input of the outer logarithm must be positive. - Determine when \(\ln x>0\). - Treat \(\ln x\) as the inner function. - Apply the chain rule step by step.

1. The outer logarithm requires \(\ln x>0\). This occurs when \(x>1\), which also satisfies the inner logarithm’s requirement \(x>0\). Thus, \(D_g=(1, \infty)\). 2. Apply the chain rule: \(g'(x)=\frac{1}{\ln x}\cdot\frac{1}{x}=\frac{1}{x\ln x}\).

\(D_g=(1, \infty)\) \(g'(x)=\frac{1}{x\ln x}\)

- Require the logarithm’s argument to be positive. - Apply the product rule. - Use the chain rule when differentiating \(\ln(3-x)\).

1. The logarithm requires \(3-x>0\), so \(x<3\). Therefore, \(D_f=(-\infty, 3)\). 2. Apply the product rule. Since \(\frac{d}{dx}\ln(3-x)=-\frac{1}{3-x}=\frac{1}{x-3}\), the derivative is \(f'(x)=2x\ln(3-x)+\frac{x^2}{x-3}\).

\(D_f=(-\infty, 3)\) \(f'(x)=2x\ln(3-x)+\frac{x^2}{x-3}\)

- Identify the inner and outer functions in part a. - Recall the derivative of the natural logarithm. - Apply the product rule in part b. - Determine the logarithm domain separately for each part.

1. For part a, apply the chain rule: \(f'(x)=\frac{-4}{1-4x}=\frac{4}{4x-1}\). The logarithm requires \(1-4x>0\), so \(x<\frac{1}{4}\). 2. For part b, apply the product rule. Since \(\frac{d}{dx}\ln(5x)=\frac{1}{x}\), \(f'(x)=2x\ln(5x)+x^2\left(\frac{1}{x}\right)=2x\ln(5x)+x\). This expression is defined for \(x>0\).

a) \(f'(x)=\frac{4}{4x-1}\), for \(x<\frac{1}{4}\) b) \(f'(x)=2x\ln(5x)+x\), for \(x>0\)

- Apply the product rule. - Differentiate cosine. - Use the chain rule for \(\ln(x^2+1)\). - Combine the two product-rule terms.

1. Apply the product rule with \(u(x)=\cos x\) and \(v(x)=\ln(x^2+1)\). 2. Their derivatives are \(u'(x)=-\sin x\) and \(v'(x)=\frac{2x}{x^2+1}\). 3. Therefore, \(f'(x)=-\sin x\ln(x^2+1)+\frac{2x\cos x}{x^2+1}\).

\(f'(x)=\frac{2x\cos x}{x^2+1}-\sin x\ln(x^2+1)\)

- Use the fact that \(\ln1=0\). - Apply the chain rule to find the derivative. - A tangent line requires a point and a slope. - Use point-slope form.

1. To find the zero, solve \(\ln(4-x)=0\). Thus, \(4-x=1\), so \(x=3\). 2. Differentiate: \(f'(x)=-\frac{1}{4-x}=\frac{1}{x-4}\). At \(x=3\), \(f(3)=0\) and \(f'(3)=-1\). Therefore, the tangent line is \(y=-(x-3)=-x+3\).

Zero: \(x=3\) Tangent line: \(y=-x+3\)

- Recall the domain restriction for a logarithm. - Use the chain rule to differentiate \(g\). - At a local extremum of \(u\), what is the value of \(u'\)? - Determine which logarithm input produces an output of \(0\).

1. Statement (1) is false. If \(u(x_0)=0\), then \(\ln(u(x_0))=\ln(0)\) is undefined, so \(x_0\) is not even in the domain of \(g\). 2. By the chain rule, \(g'(x)=\frac{u'(x)}{u(x)}\). At a local extremum \(x_0\) of \(u\), \(u'(x_0)=0\). Because \(x_0\) is in the domain of \(g\), \(u(x_0)>0\), so \(g'(x_0)=0\). Statement (2) is true. 3. If \(g(x_0)=0\), then \(\ln(u(x_0))=0\). Exponentiating gives \(u(x_0)=e^0=1\). Statement (3) is true.

(1) False (2) True (3) True

- Factor the quadratic inside the logarithm. - Determine where that expression is positive. - Apply the chain rule. - Cancel common factors only after preserving the domain restriction.

1. Factor the logarithm’s argument: \(x^2+4x+4=(x+2)^2\). This expression is positive for every real \(x\) except \(x=-2\). Thus, \(D_f=\mathbb{R}\setminus\{-2\}\). 2. Apply the chain rule: \(f'(x)=\frac{2x+4}{x^2+4x+4}\). Factor and simplify: \(f'(x)=\frac{2(x+2)}{(x+2)^2}=\frac{2}{x+2}\), for \(x\ne-2\).

\(D_f=\mathbb{R}\setminus\{-2\}\) \(f'(x)=\frac{2}{x+2}\)

- Determine whether the quadratic in part a can ever be zero or negative. - Include both logarithm and denominator restrictions in part b. - Apply the chain rule to \(\ln(4x+1)\). - Use the quotient rule in part b.

1. Complete the square in part a: \(x^2+2x+2=(x+1)^2+1>0\) for every real \(x\). Therefore, \(D_h=\mathbb{R}\). Applying the chain rule, \(h'(x)=\frac{2x+2}{x^2+2x+2}\). 2. For part b, require \(4x+1>0\), so \(x>-\frac{1}{4}\), and also require \(x\ne0\). Thus, \(D_k=\left(-\frac{1}{4}, 0\right)\cup(0, \infty)\). Using the quotient rule, \(k'(x)=\frac{\frac{4x}{4x+1}-\ln(4x+1)}{x^2}\). Combining the numerator gives \(k'(x)=\frac{4x-(4x+1)\ln(4x+1)}{x^2(4x+1)}\).

a) \(D_h=\mathbb{R}\), \(h'(x)=\frac{2x+2}{x^2+2x+2}\) b) \(D_k=\left(-\frac{1}{4}, 0\right)\cup(0, \infty)\), \(k'(x)=\frac{4x-(4x+1)\ln(4x+1)}{x^2(4x+1)}\)

- Identify the outermost operation in the function. - Use the chain rule for each composite factor. - What is the derivative of \(\cos(kx)\)? - Keep track of the negative sign when combining the two product-rule terms.

1. Let \(u(x) = (x^3-2)^5\) and \(v(x) = \cos(4x)\). 2. By the chain rule, \(u'(x) = 5(x^3-2)^4(3x^2) = 15x^2(x^3-2)^4\), and \(v'(x) = -4\sin(4x)\). 3. Apply the product rule: \(f'(x) = 15x^2(x^3-2)^4\cos(4x) - 4(x^3-2)^5\sin(4x)\).

\(f'(x) = 15x^2(x^3-2)^4\cos(4x) - 4(x^3-2)^5\sin(4x)\)

- Separate the function into two terms and differentiate each one. - The first term requires both the product rule and the chain rule. - Identify the inner function in \(\sin(x^2+1)\). - Keep the constant factor and the inner derivative when differentiating the power in the second term.

1. Differentiate the two terms separately. 2. For \(3x^2\sin(x^2+1)\), use the product rule and the chain rule: \(\frac{d}{dx}[3x^2\sin(x^2+1)] = 6x\sin(x^2+1) + 3x^2\cos(x^2+1)(2x)\) \(= 6x\sin(x^2+1) + 6x^3\cos(x^2+1)\). 3. For the second term, use the chain rule: \(\frac{d}{dx}\left[\frac{1}{4}(2x-5)^8\right] = \frac{1}{4}\cdot 8(2x-5)^7\cdot 2 = 4(2x-5)^7\). 4. Include the subtraction from the original function: \(f'(x) = 6x\sin(x^2+1) + 6x^3\cos(x^2+1) - 4(2x-5)^7\).

\(f'(x) = 6x\sin(x^2+1) + 6x^3\cos(x^2+1) - 4(2x-5)^7\)

- Replacing \(x\) by \(-x\) reflects a graph across the y-axis. - Apply the chain rule carefully to \(e^{-x}\). - Set the two exponential expressions equal for the intersection claim. - A horizontal tangent requires the derivative of the relevant function to be zero.

1. Since \(g(x)=f(-x)\), the graph of \(g\) is the reflection of the graph of \(f\) across the y-axis. Statement (1) is true. 2. The derivatives are \(f'(x)=e^x=f(x)\) and, by the chain rule, \(g'(x)=-e^{-x}=-g(x)\). Statement (2) is true. 3. Solving \(e^x=e^{-x}\) gives \(e^{2x}=1\), so \(x=0\). Statement (3) is true. 4. Since \(s'(x)=e^x+e^{-x}>0\) for every real \(x\), in particular \(s'(0)\ne0\). Statement (4) is false. 5. \(f(x)g(x)=e^xe^{-x}=1\). Statement (5) is true.

(1) True (2) True (3) True (4) False (5) True

- Use the chain rule to obtain the derivative and study its sign. - Check the relevant end behavior for the asymptote claim. - For the zero, isolate the exponential expression before using logarithms. - Determine which values the positive exponential factor can take before describing the derivative's range.

1. By the chain rule, \(h'(x)=-0.5e^{0.5x}<0\) for every real \(x\), so statement (1) is true. 2. As \(x\to-\infty\), \(e^{0.5x}\to0\), so \(h(x)\to2\). The horizontal asymptote is \(y=2\), so statement (2) is false. 3. \(h'(0)=-0.5e^0=-0.5\), so statement (3) is true. 4. Solving \(2-e^{0.5x}=0\) gives \(e^{0.5x}=2\), so \(x=2\ln2=\ln4\). Statement (4) is true. 5. Since \(e^{0.5x}\) takes every positive value, \(-0.5e^{0.5x}\) takes every negative value and never equals \(0\). Statement (5) is true.

(1) True (2) False (3) True (4) True (5) True

- Substitute the coordinates of the given point into the function. - Rewrite the square root as a power and apply the chain rule. - The derivative gives the tangent slope. - Use point-slope form for the tangent line.

1. Substitute \(A(2, 4)\): \(4=\sqrt{k-4}\). Squaring gives \(16=k-4\), so \(k=20\). 2. For \(f_{20}(x)=\sqrt{20-x^2}\), the chain rule gives \(f_{20}'(x)=-\frac{x}{\sqrt{20-x^2}}\). 3. The slope at \(x=2\) is \(f_{20}'(2)=-\frac{2}{4}=-\frac{1}{2}\). 4. Using point-slope form through \((2, 4)\), \(y-4=-\frac{1}{2}(x-2)\). Therefore, \(y=-\frac{1}{2}x+5\).

a) \(k=20\) b) \(y=-\frac{1}{2}x+5\)

- Recall the derivatives of sine and cosine. - For a trigonometric function with a linear input, multiply by the derivative of the input. - Track the negative sign when differentiating cosine. - In part c, differentiate each term separately and check the square-root domain.

1. For part a, apply the chain rule: \(f'(x)=-\sin(5x+2)\cdot5=-5\sin(5x+2)\). This derivative is defined for all real \(x\). 2. For part b, apply the constant multiple rule and chain rule: \(f'(x)=-2\cdot0.4\cos(0.4x)=-0.8\cos(0.4x)\). This derivative is defined for all real \(x\). 3. For part c, differentiate term by term: \(f'(x)=2x^3-\frac{1}{2\sqrt{x}}+\cos x\). This derivative is defined for \(x>0\).

a) \(f'(x)=-5\sin(5x+2)\), for all real \(x\) b) \(f'(x)=-0.8\cos(0.4x)\), for all real \(x\) c) \(f'(x)=2x^3-\frac{1}{2\sqrt{x}}+\cos x\), for \(x>0\)

- Identify the inner and outer functions. - Apply the chain rule to differentiate the composition. - Use the exact sine value for the special angle in part 2.

1. The outer function is cosine, and the inner function is \(x^2\). By the chain rule, \(f'(x)=-\sin(x^2)(2x)=-2x\sin(x^2)\). 2. Substitute \(x=\sqrt{\frac{\pi}{3}}\): \(f'\left(\sqrt{\frac{\pi}{3}}\right)=-2\sqrt{\frac{\pi}{3}}\sin\left(\frac{\pi}{3}\right)\). Since \(\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}\), \(f'\left(\sqrt{\frac{\pi}{3}}\right)=-2 \cdot \sqrt{\frac{\pi}{3}} \cdot \left(\frac{\sqrt{3}}{2}\right)=-\sqrt{\pi}\).

1) \(f'(x)=-2x\sin(x^2)\) 2) The exact slope is \(-\sqrt{\pi}\).

- A horizontal tangent has slope zero. - Identify which operation acts on \(x\) first and which acts on the result. - Use the unit circle or known zeros of sine and cosine.

1. One decomposition is \(v(x)=\cos(x)\) and \(u(t)=t^2\). 2. By the chain rule, \(g'(x)=2\cos(x)(-\sin(x))=-2\sin(x)\cos(x)\). Equivalently, \(g'(x)=-\sin(2x)\). 3. A horizontal tangent occurs where \(g'(x)=0\). Thus, \(-2\sin(x)\cos(x)=0\), so either \(\sin(x)=0\) or \(\cos(x)=0\). On \([0, \pi]\), these equations give \(x=0\), \(x=\frac{\pi}{2}\), and \(x=\pi\).

1) \(v(x)=\cos(x)\) and \(u(t)=t^2\) 2) \(g'(x)=-2\sin(x)\cos(x)=-\sin(2x)\) 3) \(x=0\), \(x=\frac{\pi}{2}\), and \(x=\pi\)

- Evaluate the composition from the inside out. - Apply the chain rule once for each layer of the composition. - The derivative is the product of the derivative factors from the outermost function to the innermost function. - A logarithm property can provide a useful check.

1. Work from the inside out: \(v(w(x))=\sqrt{x^2+5}\), so \(f(x)=\ln\left(\sqrt{x^2+5}\right)\). Equivalently, \(f(x)=\frac{1}{2}\ln(x^2+5)\). 2. Applying the chain rule to all three layers gives \(f'(x)=\frac{1}{\sqrt{x^2+5}}\left(\frac{1}{2\sqrt{x^2+5}}\right)(2x)\) \(=\frac{x}{x^2+5}\).

1) \(f(x)=\ln\left(\sqrt{x^2+5}\right)=\frac{1}{2}\ln(x^2+5)\) 2) \(f'(x)=\frac{x}{x^2+5}\)

- List the operations in the order in which they are performed on an input. - Recall that \(\cos^4(z)\) means \((\cos(z))^4\). - Remember the negative sign in the derivative of cosine. - Combine the constant factors after applying the chain rule.

1. One decomposition is \(w(x)=3x-1\), \(v(t)=\cos(t)\), and \(u(t)=t^4\). 2. Apply the chain rule through all three layers: \(h'(x)=4\cos^3(3x-1)[-\sin(3x-1)]\cdot 3\) \(=-12\cos^3(3x-1)\sin(3x-1)\).

1) \(w(x)=3x-1\), \(v(t)=\cos(t)\), and \(u(t)=t^4\) 2) \(h'(x)=-12\cos^3(3x-1)\sin(3x-1)\)

- Build the composition from the innermost function outward. - Substitute the entire output of one function into the next function. - Apply the chain rule once for each nested layer. - Recall the derivatives of cosine and the exponential function.

1. First, \(q(r(x))=\cos^2(x)-4\). Therefore, \(f(x)=e^{\cos^2(x)-4}\). 2. Differentiate the exponent using the chain rule: \(\frac{d}{dx}[\cos^2(x)-4]=2\cos(x)(-\sin(x))=-2\sin(x)\cos(x)\). 3. Apply the chain rule to the exponential function: \(f'(x)=-2\sin(x)\cos(x)e^{\cos^2(x)-4}\). Since \(2\sin(x)\cos(x)=\sin(2x)\), this can also be written as \(f'(x)=-\sin(2x)e^{\cos^2(x)-4}\).

a) \(f(x)=e^{\cos^2(x)-4}\) b) \(f'(x)=-2\sin(x)\cos(x)e^{\cos^2(x)-4}=-\sin(2x)e^{\cos^2(x)-4}\)

- Describe the function as three operations performed in sequence. - Recall the derivatives of the square-root and exponential functions. - Simplify after multiplying the three derivative factors. - Rewrite the expression using exponent rules to check your result.

1. One decomposition is \(u(t)=\sqrt{t}\), \(v(t)=e^t\), and \(w(x)=x^2+4\). 2. Their derivatives are \(u'(t)=\frac{1}{2\sqrt{t}}\), \(v'(t)=e^t\), and \(w'(x)=2x\). 3. By the chain rule, \(g'(x)=\frac{1}{2\sqrt{e^{x^2+4}}}e^{x^2+4}(2x)\) \(=x\sqrt{e^{x^2+4}}\). 4. As a check, \(g(x)=e^{\frac{1}{2}x^2+2}\), whose derivative is \(g'(x)=xe^{\frac{1}{2}x^2+2}\), the same result.

\(g'(x)=x\sqrt{e^{x^2+4}}=xe^{\frac{1}{2}x^2+2}\)

- In the quotient rule, the derivative of the constant numerator is zero. - Rewrite \(\frac{1}{z}\) as \(z^{-1}\) for the chain-rule method. - Track the negative sign and the new exponent carefully.

1. Using the quotient rule with numerator \(5\) and denominator \(x^2+3\), \(f'(x)=\frac{0(x^2+3)-5(2x)}{(x^2+3)^2}\) \(=-\frac{10x}{(x^2+3)^2}\). 2. Rewrite the function as \(f(x)=5(x^2+3)^{-1}\). Using the chain rule, \(f'(x)=5[-(x^2+3)^{-2}](2x)\) \(=-\frac{10x}{(x^2+3)^2}\). The two methods agree.

\(f'(x)=-\frac{10x}{(x^2+3)^2}\)

- Use the degree rule for polynomial composition. - Differentiating a nonconstant polynomial lowers its degree by \(1\). - The degree of a product is the sum of the degrees. - Factor the equation obtained by equating the two degrees.

1. The degree of a polynomial composition is the product of the degrees, so \(\deg(f)=nm\). 2. A nonconstant polynomial of degree \(nm\) has a derivative of degree \(nm-1\). This also follows from the chain rule: \(f'(x)=u'(v(x))v'(x)\), whose factors have degrees \((n-1)m\) and \(m-1\), for a total of \(nm-1\). 3. The product \(h(x)=u(x)v(x)\) has degree \(n+m\). Equating the degrees gives \(nm=n+m\), or \((n-1)(m-1)=1\). Since \(n\) and \(m\) are positive integers, the only solution is \(n=m=2\).

1) \(\deg(f)=nm\) 2) \(\deg(f')=nm-1\) 3) \(n=2\) and \(m=2\)

- Apply the chain rule to \(f=p\circ q\). - Find the degree of each factor in \(f'(x)=p'(q(x))q'(x)\). - List the ordered factor pairs that satisfy the restrictions.

1. By the chain rule, \(f'(x)=p'(q(x))q'(x)\). The factors have degrees \((n-1)m\) and \(m-1\), so \(\deg(f')=(n-1)m+(m-1)=nm-1\). 2. Since \(f'\) has degree \(11\), \(nm-1=11\), so \(nm=12\). 3. The ordered factor pairs of \(12\) with both entries at least \(2\) are \((2, 6)\), \((3, 4)\), \((4, 3)\), and \((6, 2)\).

\((2, 6)\), \((3, 4)\), \((4, 3)\), and \((6, 2)\)

- Use the chain rule for \(f\). - A tangent line needs a point and a derivative value. - Equal exponential expressions with the same base have equal exponents. - Compare the two derivative values at the intersection input.

1. By the chain rule, \(f'(x)=2e^{2x-4}\). At \(x=2\), \(f(2)=1\) and \(f'(2)=2\). Thus, the tangent line is \(y-1=2(x-2)\), or \(y=2x-3\). 2. At an intersection, \(e^{2x-4}=e^x\). Since the bases are equal and the exponential function is one-to-one, \(2x-4=x\), so \(x=4\). 3. Since \(g'(x)=e^x\), \(g'(4)=e^4\). Also, \(f'(4)=2e^{2\cdot4-4}=2e^4\). Therefore, \(f'(4)=2g'(4)\).

a) \(f'(x)=2e^{2x-4}\); tangent line: \(y=2x-3\) b) The x-coordinate of \(S\) is \(4\). c) \(f'(4)=2e^4=2g'(4)\).

- First find the derivative using the chain rule. - The slope at a point is the value of the first derivative there. - Use radians for the trigonometric inputs. - Treat the power in part c as the outer function.

1. For a), \(f'(x)=\frac{x}{\sqrt{x^2+9}}\), so \(f'(4)=\frac{4}{\sqrt{25}}=\frac{4}{5}=0.8\). 2. For b), \(f'(x)=2\cos(2x)\), so \(f'\left(\frac{\pi}{6}\right)=2\cos\left(\frac{\pi}{3}\right)=1\). 3. For c), \(f'(x)=3(\cos(x)+1)^2(-\sin(x))\), so \(f'\left(\frac{\pi}{2}\right)=3\cdot 1^2\cdot(-1)=-3\).

a) \(f'(4)=0.8\) b) \(f'\left(\frac{\pi}{6}\right)=1\) c) \(f'\left(\frac{\pi}{2}\right)=-3\)

- Identify the inner and outer functions. - Expand the cube by multiplying or using the binomial theorem. - Expand or factor one derivative expression to compare the two forms.

1. Using the chain rule, \(f'(x)=3(0.5x^2-2)^2(x)=3x(0.5x^2-2)^2\). 2. Expanding first gives \(f(x)=0.125x^6-1.5x^4+6x^2-8\). Therefore, \(f'(x)=0.75x^5-6x^3+12x\). 3. Expand the chain-rule result: \(3x(0.5x^2-2)^2=3x(0.25x^4-2x^2+4)\) \(=0.75x^5-6x^3+12x\). Thus, the results are equivalent.

1) \(f'(x)=3x(0.5x^2-2)^2\) 2) \(f'(x)=0.75x^5-6x^3+12x\) 3) Expanding the first expression gives the second expression.

- Rewrite radicals as fractional powers when useful. - A fraction with a constant numerator can be rewritten using a negative exponent. - Differentiate the terms of a sum separately.

1. For a), \(g'(x)=\frac{1}{2\sqrt{x^2+2x+5}}(2x+2)\) \(=\frac{x+1}{\sqrt{x^2+2x+5}}\). This derivative is defined for all real \(x\). 2. For b), rewrite \(h(x)=10(3x-1)^{-4}\). Then \(h'(x)=10[-4(3x-1)^{-5}]\cdot 3\) \(=-\frac{120}{(3x-1)^5}\), which is defined for \(x\ne\frac{1}{3}\). 3. For c), differentiate each term: \(k'(x)=\frac{1}{2\sqrt{2x}}\cdot 2+2(x^3-1)(3x^2)\) \(=\frac{1}{\sqrt{2x}}+6x^2(x^3-1)\). This derivative is defined for \(x>0\).

a) \(g'(x)=\frac{x+1}{\sqrt{x^2+2x+5}}\), for all real \(x\) b) \(h'(x)=-\frac{120}{(3x-1)^5}\), for \(x\ne\frac{1}{3}\) c) \(k'(x)=\frac{1}{\sqrt{2x}}+6x^2(x^3-1)\), for \(x>0\)

- Use the product rule for the two main factors. - Each factor is also a composite function. - Apply the chain rule carefully to \(\sin^3(2x)\). - Track the sign when differentiating cosine.

1. Let \(u(x)=\cos(x^2)\) and \(v(x)=\sin^3(2x)\). 2. By the chain rule, \(u'(x)=-2x\sin(x^2)\). Also, \(v'(x)=3\sin^2(2x)\cos(2x)\cdot 2=6\sin^2(2x)\cos(2x)\). 3. Apply the product rule: \(f'(x)=-2x\sin(x^2)\sin^3(2x)+6\cos(x^2)\sin^2(2x)\cos(2x)\).

\(f'(x)=-2x\sin(x^2)\sin^3(2x)+6\cos(x^2)\sin^2(2x)\cos(2x)\)

- Treat the square root as the outer function. - Use the quotient rule on the expression under the radical. - Multiply the outer derivative by the inner derivative. - Track signs carefully when expanding the quotient-rule numerator.

1. Let \(r(x)=\frac{3x+2}{x^2+1}\). The outer function is \(\sqrt{r}\), whose derivative is \(\frac{1}{2\sqrt{r}}\). 2. By the quotient rule, \(r'(x)=\frac{3(x^2+1)-(3x+2)(2x)}{(x^2+1)^2}\) \(=\frac{-3x^2-4x+3}{(x^2+1)^2}\). 3. Apply the chain rule: \(f'(x)=\frac{1}{2\sqrt{\frac{3x+2}{x^2+1}}}\left(\frac{-3x^2-4x+3}{(x^2+1)^2}\right)\) \(=\frac{-3x^2-4x+3}{2(x^2+1)^2\sqrt{\frac{3x+2}{x^2+1}}}\). The derivative is defined for \(x>-\frac{2}{3}\).

\(f'(x)=\frac{-3x^2-4x+3}{2(x^2+1)^2\sqrt{\frac{3x+2}{x^2+1}}}\), for \(x>-\frac{2}{3}\)

- Use the quotient rule for the outer structure. - Apply the chain rule to the radical denominator. - Combine the numerator over a common radical denominator. - Use exponent rules to simplify the final denominator.

1. Let \(u(x)=kx\) and \(v(x)=\sqrt{x^2+1}\). Then \(u'(x)=k\) and \(v'(x)=\frac{x}{\sqrt{x^2+1}}\). 2. By the quotient rule, \(f_k'(x)=\frac{k\sqrt{x^2+1}-kx\frac{x}{\sqrt{x^2+1}}}{x^2+1}\). 3. Combine the numerator: \(f_k'(x)=\frac{k(x^2+1)-kx^2}{(x^2+1)\sqrt{x^2+1}}=\frac{k}{(x^2+1)^{\frac{3}{2}}}\).

\(f_k'(x)=\frac{k}{(x^2+1)^{\frac{3}{2}}}\)

- Use the product rule for the two main factors. - Apply the chain rule to the cubic factor. - Look for a common power of \(3x+1\) before expanding. - Keep the final answer in factored form.

1. Let \(u(x)=x^2-4\) and \(v(x)=(3x+1)^3\). 2. Then \(u'(x)=2x\) and, by the chain rule, \(v'(x)=9(3x+1)^2\). 3. Apply the product rule: \(g'(x)=2x(3x+1)^3+9(x^2-4)(3x+1)^2\). 4. Factor out \((3x+1)^2\): \(g'(x)=(3x+1)^2[2x(3x+1)+9(x^2-4)]\). 5. Simplify the remaining factor: \(g'(x)=(3x+1)^2(15x^2+2x-36)\).

\(g'(x)=(3x+1)^2(15x^2+2x-36)\)

- What derivative value corresponds to a horizontal tangent? - Use both the product rule and the chain rule. - Include the derivative of the inner expression \(c-x\). - Substitute \(x=\frac{c}{2}\) before doing unnecessary simplification.

1. Differentiate using the product and chain rules: \(f'(x)=\sec^2(x)\tan(c-x)-\tan(x)\sec^2(c-x)\). 2. Substitute \(x=\frac{c}{2}\): \(f'\left(\frac{c}{2}\right)=\sec^2\left(\frac{c}{2}\right)\tan\left(\frac{c}{2}\right)-\tan\left(\frac{c}{2}\right)\sec^2\left(\frac{c}{2}\right)=0\). 3. Therefore, the tangent slope is \(0\), so the graph has a horizontal tangent at \(x=\frac{c}{2}\). The condition that \(f\) is defined there is equivalent to \(\cos\left(\frac{c}{2}\right)\neq0\).

\(f'\left(\frac{c}{2}\right)=0\), so the tangent is horizontal whenever \(\cos\left(\frac{c}{2}\right)\neq0\).

- Identify the outer product and the composite functions inside it. - Use the chain rule for each factor. - Differentiate the two factors separately before applying the product rule. - Factor out the largest common power of \(x^2+1\).

1. Let \(u(x)=(x^2+1)^4\) and \(v(x)=\sin(5x)\). 2. By the chain rule, \(u'(x)=8x(x^2+1)^3\) and \(v'(x)=5\cos(5x)\). 3. Apply the product rule: \(f'(x)=8x(x^2+1)^3\sin(5x)+5(x^2+1)^4\cos(5x)\). 4. Factor out \((x^2+1)^3\): \(f'(x)=(x^2+1)^3[8x\sin(5x)+5(x^2+1)\cos(5x)]\).

\(f'(x)=(x^2+1)^3[8x\sin(5x)+5(x^2+1)\cos(5x)]\)

- Find the point and slope needed for a tangent-line equation. - Use point-slope form at \(x_0=4\). - Apply the chain rule when differentiating the exponential function. - Set \(y=0\) for the x-intercept and \(x=0\) for the y-intercept. - Use the intercept distances as the legs of the right triangle.

1. Evaluate the function: \(f(4)=4e^{-1}=\frac{4}{e}\). Differentiate using the chain rule: \(f'(x)=-e^{-0.25x}\), so \(f'(4)=-\frac{1}{e}\). 2. Use point-slope form: \(y-\frac{4}{e}=-\frac{1}{e}(x-4)\). Therefore, \(t: y=-\frac{1}{e}x+\frac{8}{e}\). 3. For the y-intercept, set \(x=0\), giving \(\left(0, \frac{8}{e}\right)\). For the x-intercept, set \(y=0\), giving \(x=8\), so the intercept is \((8, 0)\). 4. The triangle has legs \(8\) and \(\frac{8}{e}\). Its area is \(A=\frac{1}{2}\cdot8\cdot\frac{8}{e}=\frac{32}{e}\approx11.77\) square units.

1. \(t: y=-\frac{1}{e}x+\frac{8}{e}\) 2. x-intercept: \((8, 0)\); y-intercept: \(\left(0, \frac{8}{e}\right)\) 3. \(A=\frac{32}{e}\approx11.77\) square units

- Work with \(x_0\) and \(k\) symbolically in part 1. - At an x-intercept, the y-coordinate is \(0\). - Factor out \(e^{kx_0}\) from the intercept equation. - Use the result from part 1 for parts 2 and 3. - The origin has x-coordinate \(0\).

1. Differentiate: \(f_k'(x)=ke^{kx}\). The tangent line at \(x_0\) is \(y=ke^{kx_0}(x-x_0)+e^{kx_0}\). 2. At the x-intercept, \(y=0\). Thus \(0=e^{kx_0}[k(x_s-x_0)+1]\). Since \(e^{kx_0}>0\), \(k(x_s-x_0)+1=0\), so \(x_s=x_0-\frac{1}{k}\). 3. Substitute \(x_0=3\) and \(x_s=1\): \(1=3-\frac{1}{k}\). Therefore, \(\frac{1}{k}=2\), so \(k=0.5\). 4. A line through the origin has x-intercept \(x_s=0\). With \(k=0.5\), \(0=x_0-\frac{1}{0.5}=x_0-2\), so \(x_0=2\).

1. \(x_s=x_0-\frac{1}{k}\) 2. \(k=0.5\) 3. \(x_0=2\)

- Parallel lines have the same slope. - Use the chain rule to differentiate the logarithmic function. - Setting the derivative equal to the line's slope produces a quadratic equation. - Use point-slope form for each tangent line.

1. Rewrite the given line as \(y=\frac{4}{5}x+2\), so its slope is \(\frac{4}{5}\). 2. Differentiate using the chain rule: \(f'(x)=\frac{2x}{x^2+1}\). 3. Set the derivative equal to the required slope: \(\frac{2x}{x^2+1}=\frac{4}{5}\). This gives \(4x^2-10x+4=0\), or \(2x^2-5x+2=0\). 4. Factor: \((2x-1)(x-2)=0\). Thus \(x=\frac{1}{2}\) or \(x=2\). 5. The corresponding points are \(P_1\left(\frac{1}{2}, \ln\left(\frac{5}{4}\right)\right)\) and \(P_2(2, \ln 5)\). 6. Using slope \(\frac{4}{5}\), the tangent lines are \(y-\ln\left(\frac{5}{4}\right)=\frac{4}{5}\left(x-\frac{1}{2}\right)\) and \(y-\ln 5=\frac{4}{5}(x-2)\).

Points: \(P_1\left(\frac{1}{2}, \ln\left(\frac{5}{4}\right)\right)\) and \(P_2(2, \ln 5)\) Tangent lines: \(y=\frac{4}{5}x-\frac{2}{5}+\ln\left(\frac{5}{4}\right)\) and \(y=\frac{4}{5}x-\frac{8}{5}+\ln 5\)

- Start with the general form of a cubic polynomial. - Tangency at \(x=0\) gives one condition for the value and one for the slope. - A horizontal tangent means the derivative is zero. - Use all conditions to form and solve a linear system.

1. Write \(p(x)=ax^3+bx^2+cx+d\), so \(p'(x)=3ax^2+2bx+c\). 2. Tangency at \(x=0\) requires equal function values and slopes. Since \(f(0)=4\), \(d=4\). 3. Differentiate \(f\): \(f'(x)=-2e^{-0.5x}\), so \(f'(0)=-2\). Therefore, \(c=-2\). 4. The horizontal tangent at \(x=-1\) gives \(p'(-1)=3a-2b-2=0\), so \(3a-2b=2\). 5. Since \(p(1)=2\), \(a+b-2+4=2\), so \(a+b=0\). 6. Solving the system gives \(a=\frac{2}{5}\) and \(b=-\frac{2}{5}\). 7. Therefore, \(p(x)=\frac{2}{5}x^3-\frac{2}{5}x^2-2x+4\).

\(p(x)=\frac{2}{5}x^3-\frac{2}{5}x^2-2x+4\)

- A line through the origin has equation \(y=mx\). - At the point of tangency, the line's slope equals the derivative. - Use the chain rule to differentiate \(\ln(x^2)\). - Determine how many solutions the equation for \(u\) has.

1. Differentiate using the chain rule: \(f'(x)=\frac{2}{x}\). 2. Let the point of tangency have x-coordinate \(u\neq0\). A tangent line through the origin has slope \(\frac{2}{u}\), so its equation is \(y=\frac{2}{u}x\). 3. The point \((u, \ln(u^2))\) must lie on this line. Therefore, \(\ln(u^2)=\frac{2}{u}u=2\). 4. Solving \(\ln(u^2)=2\) gives \(u^2=e^2\), so \(u=e\) or \(u=-e\). These are the only two solutions, so exactly two such tangent lines exist. 5. Their slopes are \(\frac{2}{e}\) and \(-\frac{2}{e}\). Thus the tangent lines are \(y=\frac{2}{e}x\) and \(y=-\frac{2}{e}x\).

a) The two points of tangency occur at \(u=e\) and \(u=-e\). b) \(y=\frac{2}{e}x\) and \(y=-\frac{2}{e}x\)

- Rewrite the function so the coefficient and vertical shift are easy to identify. - What happens to the exponential term as time increases without bound? - Use the chain rule to differentiate the exponential expression. - In a time-dependent process, what does the first derivative represent?

1. Rewrite the function as \(U(t)=-15e^{-0.4t}+15\). Starting with \(g(t)=e^{-0.4t}\), reflect the graph across the \(t\)-axis, stretch it vertically by a factor of \(15\), and shift it up \(15\) units. 2. Since \(e^{-0.4t}\to 0\) as \(t\to\infty\), \(U(t)\to 15\). The horizontal asymptote is \(U=15\). It represents the limiting voltage of the capacitor. 3. Differentiate using the chain rule: \(U'(t)=15\left(0.4e^{-0.4t}\right)=6e^{-0.4t}\). 4. At \(t=2\), \(U'(2)=6e^{-0.8}\approx 2.70\,\text{V/s}\). At that instant, the voltage is increasing at about \(2.70\) volts per second.

a) Reflect across the \(t\)-axis, stretch vertically by a factor of \(15\), and shift up \(15\) units. b) The horizontal asymptote is \(U=15\). The capacitor voltage approaches \(15\,\text{V}\). c) \(U'(t)=6e^{-0.4t}\), and \(U'(2)\approx 2.70\,\text{V/s}\).

- Convert each graph condition into an equation involving \(f\) or \(f'\). - Differentiate the general function before using the slope condition. - Use exact sine and cosine values at \(0\), \(\frac{\pi}{2}\), and \(\pi\). - Solve the resulting linear system for the three parameters.

1. Translate the conditions into equations: \(f(0)=4\), \(f\left(\frac{\pi}{2}\right)=1\), and \(f'(0)=2\). 2. Differentiate: \(f'(x)=a\cos x-2b\sin(2x)\). From \(f'(0)=2\), \(a=2\). 3. From \(f(0)=4\), \(b+c=4\). 4. From \(f\left(\frac{\pi}{2}\right)=1\), \(2-b+c=1\), so \(-b+c=-1\). 5. Solve the system \(b+c=4\) and \(-b+c=-1\). This gives \(b=\frac{5}{2}\) and \(c=\frac{3}{2}\).

\(a=2\), \(b=\frac{5}{2}\), and \(c=\frac{3}{2}\) Therefore, \(f(x)=2\sin x+\frac{5}{2}\cos(2x)+\frac{3}{2}\).

- Evaluate a composition from the inside out. - Use the chain rule for a function of the form \(f(g(x))\). - A horizontal tangent occurs where the derivative is zero. - Use the zero-product property on the two derivative factors.

1. From the graph, \(f(2)=3\) and \(f(4)=4\). Therefore, \(h(1)=f(1^2+3)=f(4)=4\). 2. By the chain rule, \(h'(x)=2x f'(x^2+3)\). 3. A horizontal tangent occurs where \(2x f'(x^2+3)=0\). One possibility is \(2x=0\), which gives \(x=0\). The graph of \(f\) has a horizontal tangent only when its input is \(4\), so the other possibility is \(x^2+3=4\). Thus, \(x^2=1\), giving \(x=-1\) or \(x=1\). Therefore, the graph of \(h\) has horizontal tangents at \(x=-1\), \(x=0\), and \(x=1\).

a) \(f(2)=3\), \(f(4)=4\), and \(h(1)=4\) b) \(h'(x)=2x f'(x^2+3)\); the horizontal tangents occur at \(x=-1\), \(x=0\), and \(x=1\).

- Use the horizontal asymptote to identify \(c\). - Use the y-intercept to find \(a\). - Substitute the graph's zero to find \(b\). - Differentiate using the chain rule and evaluate at \(0\) and \(1\).

1. The graph approaches the horizontal asymptote \(y=4\) as \(x\to-\infty\), so \(c=4\). The y-intercept is \((0, 2)\), so \(a+4=2\) and \(a=-2\). 2. The graph has a zero at \(x=1\). Thus, \(-2e^b+4=0\), so \(e^b=2\) and \(b=\ln2\). 3. Therefore, \(f(x)=-2e^{(\ln2)x}+4\), and \(f'(x)=-2\ln2\,e^{(\ln2)x}\). Hence, \(f'(0)=-2\ln2\) and \(f'(1)=-4\ln2=2f'(0)\).

a) \(c=4\), \(a=-2\) b) \(b=\ln2\) c) \(f'(0)=-2\ln2\) and \(f'(1)=-4\ln2=2f'(0)\)

- Evaluate \(f(2)\) to find the point of tangency. - Rewrite the function with a negative exponent and apply the chain rule. - The derivative value at \(x=2\) is the tangent slope. - Use point-slope form for the tangent line. - Use the absolute value of the slope when finding the acute angle with the x-axis.

1. Find the point of tangency: \(f(2)=\frac{8}{2^2+4}=1\), so \(P=(2,1)\). 2. Differentiate using the chain rule: \(f'(x)=-\frac{16x}{(x^2+4)^2}\). 3. The tangent slope is \(f'(2)=-\frac{32}{64}=-0.5\). 4. Use point-slope form: \(y-1=-0.5(x-2)\), so \(y=-0.5x+2\). 5. For the acute angle \(\beta\), use \(\tan(\beta)=|-0.5|\). Thus, \(\beta=\arctan(0.5)\approx26.6^\circ\).

a) \(t: y=-0.5x+2\) b) \(\beta\approx26.6^\circ\)

- A logarithm requires a positive argument. - Apply the chain rule to \(\ln(u(x))\). - Use the monotonicity and zero of \(f\) to determine where \(f(x)>0\). - Express \(g'\) in terms of \(f\) and \(f'\), then simplify the equation. - Read the required function value from the graph.

1. For part a, require \(3x-5>0\), so \(D_h=\left(\frac{5}{3}, \infty\right)\). The zero satisfies \(3x-5=1\), giving \(x=2\). By the chain rule, \(h'(x)=\frac{3}{3x-5}\). 2. Since \(f\) is strictly increasing and has its only zero at \(-2\), \(f(x)>0\) exactly when \(x>-2\). Therefore, \(D_g=(-2, \infty)\). Also, \(g'(x)=\frac{f'(x)}{f(x)}\). The equation \(\frac{f'(x)}{f(x)}=\frac{1}{2}f'(x)\) can be divided by \(f'(x)>0\), giving \(f(x)=2\). From the graph, \(f(x)=2\) at \(x=2\).

a) \(D_h=\left(\frac{5}{3}, \infty\right)\), zero at \(x=2\), and \(h'(x)=\frac{3}{3x-5}\) b) 1. \(D_g=(-2, \infty)\) 2. \(x=2\)

- Apply the chain rule to \(\sin(kx)\). - Compare the maximum vertical distances from the midline. - The greatest slope of the original function is the maximum value of its derivative.

1. By the chain rule, \(h^{\prime}(x)=2\cos(2x)\). 2. The function \(h(x)=\sin(2x)\) has amplitude \(1\), while \(h^{\prime}(x)=2\cos(2x)\) has amplitude \(2\). Therefore, graph p represents \(h\), and graph q represents \(h^{\prime}\). 3. The greatest slope of \(h\) is the maximum value of \(h^{\prime}(x)\). Since \(2\cos(2x)\le2\), the maximum slope is \(2\). 4. On \([0,\pi]\), \(\cos(2x)=1\) at \(x=0\) and \(x=\pi\).

a) \(h^{\prime}(x)=2\cos(2x)\) b) Graph p is \(h\); graph q is \(h^{\prime}\). c) The maximum slope is \(2\), attained at \(x=0\) and \(x=\pi\).

- Evaluate the inner function first, then use its output as the input of the outer function. - Apply the chain rule \(f^{\prime}(x)=u^{\prime}(v(x))v^{\prime}(x)\). - Read each required slope from the line or tangent shown. - Use point-slope form to write the tangent equation.

1. From the graph, \(v(1)=2\), and then \(u(2)=-1\). Therefore, \(f(1)=u(v(1))=-1\). 2. By the chain rule, \(f^{\prime}(1)=u^{\prime}(v(1))v^{\prime}(1)=u^{\prime}(2)v^{\prime}(1)\). 3. The graph of \(v\) is a line with slope \(1\), so \(v^{\prime}(1)=1\). The displayed tangent to \(u\) at \(x=2\) has slope \(1\), so \(u^{\prime}(2)=1\). Thus, \(f^{\prime}(1)=1\). 4. The tangent through \((1, -1)\) with slope \(1\) is \(y+1=x-1\), or \(y=x-2\). 5. Its intercepts are \((2, 0)\) and \((0, -2)\). The triangle has area \(\frac{1}{2}\cdot2\cdot2=2\) square units.

a) \(f(1)=-1\) and \(f^{\prime}(1)=1\) b) \(2\) square units

- A horizontal tangent occurs where the derivative is zero. - Apply the chain rule to the square of a function. - Look for both zeros of \(f\) and points where the graph of \(f\) has a horizontal tangent.

1. By the chain rule, \(g'(x)=2f(x)f'(x)\). 2. A horizontal tangent occurs where \(g'(x)=0\). By the zero-product property, this happens when \(f(x)=0\) or \(f'(x)=0\). 3. The graph shows that \(f(x)=0\) at \(x=1\) and \(x=5\). 4. The graph of \(f\) has a horizontal tangent at its vertex, \(x=3\), so \(f'(3)=0\). Therefore, the graph of \(g\) has horizontal tangents at \(x=1\), \(x=3\), and \(x=5\).

The horizontal tangents occur at \(x=1\), \(x=3\), and \(x=5\).

- Evaluate the inner function before the outer function. - Find another x-value where the inner function has the same output. - Use the chain rule to find the tangent slope. - Check the tangent’s intercepts before calculating the triangle’s area.

1. From the graph, \(v(1)=1\), so \(f(1)=u(v(1))=u(1)=1\). 2. The graph of \(v\) also has \(v(3)=1\), so \(f(3)=u(1)=1\). 3. By the chain rule, \(f^{\prime}(1)=u^{\prime}(v(1))v^{\prime}(1)=u^{\prime}(1)v^{\prime}(1)\). 4. The tangent to \(u\) has slope \(\frac{1}{2}\), and the tangent to \(v\) at \(x=1\) has slope \(-1\). Thus, \(f^{\prime}(1)=\frac{1}{2}\cdot(-1)=-0.5\). 5. The tangent through \((1, 1)\) is \(y-1=-0.5(x-1)\), or \(y=-0.5x+1.5\). 6. Its intercepts are \((3, 0)\) and \((0, 1.5)\), so the triangle’s area is \(\frac{1}{2}\cdot3\cdot1.5=2.25\) square units.

a) \(f(1)=1\); another x-value is \(x=3\). b) The tangent is \(y=-0.5x+1.5\), and the area is \(2.25\) square units.

- Evaluate the inner graph first. - Find all inputs of \(u\) that produce the same output. - Use the chain rule for the tangent slope. - Find the tangent line intercepts before computing the triangle area.

1. The graph gives \(v(1)=2\) and \(u(2)=0\), so \(f(1)=0\). 2. The zeros of \(u\) occur at inputs \(0\) and \(2\). Besides \(v(1)=2\), the graph gives \(v(3)=0\). Therefore, the second value is \(x=3\). 3. By the chain rule, \(f'(x)=u'(v(x))v'(x)\). From the graphs, \(v'(x)=-1\) and \(u'(2)=2\), so \(f'(1)=-2\). 4. The tangent line through \((1, 0)\) is \(y=-2(x-1)=-2x+2\). Its intercepts are \((1, 0)\) and \((0, 2)\). The triangle area is \(\frac{1}{2}\cdot1\cdot2=1\).

a) \(f(1)=0\); the second value is \(x=3\). b) \(1\) square unit

- Use point-slope form at \(x=a\). - At an x-intercept, the y-coordinate is \(0\). - Divide by the nonzero exponential factor. - The horizontal distance is the absolute difference of the x-coordinates.

1. Differentiate: \(f'(x)=0.5e^{0.5x}\). 2. The tangent line at \(x=a\) is \(t: y=0.5e^{0.5a}(x-a)+e^{0.5a}\). 3. At the x-intercept \(Q(x_Q, 0)\), \(0=0.5e^{0.5a}(x_Q-a)+e^{0.5a}\). Since \(e^{0.5a}>0\), divide by it to get \(0=0.5(x_Q-a)+1\). 4. Solving gives \(x_Q=a-2\). 5. The horizontal distance is \(|a-x_Q|=|a-(a-2)|=2\), independent of \(a\). In the figure, \(a=2\), so \(P=(2,e)\) and \(Q=(0,0)\), which are \(2\) units apart horizontally.

a) \(t: y=0.5e^{0.5a}(x-a)+e^{0.5a}\) b) \(x_Q=a-2\) c) \(|a-(a-2)|=2\), so the subtangent is always \(2\).

- Evaluate the function at \(x=0\). - A change in an outside multiplier produces a vertical scaling. - Use the point and slope at \(x=0\) to write the tangent line.

1. Since \(g_k(0)=k\), the y-intercepts give p: \(k=2\) and q: \(k=4\). 2. Doubling \(k\) doubles every function value, which is a vertical stretch by a factor of \(2\). 3. The derivative is \(g_k'(x)=-ke^{-x}\). At \(x=0\), the point is \((0, k)\) and the slope is \(-k\). 4. The tangent line is \(y=-kx+k\). Setting \(y=0\) gives \(x=1\), independent of \(k\).

a) p: \(k=2\); q: \(k=4\) b) Vertical stretch by a factor of \(2\) c) The tangent is \(y=-kx+k\), with x-intercept \(1\).

- A horizontal tangent gives a zero-derivative condition. - Use both the product rule and the chain rule when differentiating. - The exponential factor itself cannot make the derivative zero.

1. By the product and chain rules, \(f_k'(x)=e^{-0.5x}(1-0.5x+0.5k)\). 2. A horizontal tangent at \(x=2\) requires \(f_k'(2)=0\). 3. This gives \(e^{-1}(0.5k)=0\). Since \(e^{-1}\ne0\), \(k=0\).

\(k=0\)

- The horizontal asymptote identifies the vertical shift. - Use the y-intercept to find \(a\). - Substitute the given point and compare exponents. - Differentiate to find the tangent slope.

1. The horizontal asymptote is \(y=3\), so \(c=3\). The y-intercept is \((0, 1)\), so \(a+3=1\) and \(a=-2\). 2. Substitute \(P(2, 3-2e)\) into \(f(x)=-2e^{bx}+3\): \(3-2e=-2e^{2b}+3\). Thus, \(e=e^{2b}\), so \(b=\frac{1}{2}\). 3. The function is \(f(x)=-2e^{x/2}+3\), so \(f'(x)=-e^{x/2}\). At \(x=0\), the slope is \(-1\), and the point is \((0, 1)\). Therefore, the tangent line is \(y=-x+1\).

a) \(c=3\), \(a=-2\) b) \(b=\frac{1}{2}\) c) \(y=-x+1\)

- A point on the y-axis has x-coordinate \(0\). - Compare \(f(-x)\) with \(f(x)\). - Use the chain rule to find the tangent slope. - Apply point-slope form for the tangent line. - An x-intercept has y-coordinate \(0\).

1. Since \(f(0)=4\), the y-intercept is \((0,4)\). 2. Since \(f(-x)=4e^{-(-x)^2/2}=4e^{-x^2/2}=f(x)\), the function is even and its graph has y-axis symmetry. 3. By the chain rule, \(f'(x)=-4xe^{-x^2/2}\). At \(x=1\), the slope is \(-4e^{-1/2}\), and the point of tangency is \((1,4e^{-1/2})\). Therefore, \(y-4e^{-1/2}=-4e^{-1/2}(x-1)\), so \(t(x)=-4e^{-1/2}x+8e^{-1/2}\). 4. Set \(t(x)=0\): \(-4e^{-1/2}x+8e^{-1/2}=0\), which gives \(x=2\).

a) \((0,4)\). b) \(f(-x)=f(x)\), so the graph has y-axis symmetry. c) \(t(x)=-4e^{-1/2}x+8e^{-1/2}\). d) \(x=2\).

- What values can \(f(x)\) take according to the graph? - How does the increasing function \(e^u\) transform inequalities between exponent values? - In the chain rule, which factor determines the sign of \(g'(x)\)?

1. The quadratic function \(f\) is defined for every real number, and \(e^u\) is defined for every real input \(u\). Therefore, the domain of \(g\) is \(\mathbb{R}\). 2. From the graph, \(f\) has a maximum value of \(1\) at \(x=0\), and \(f(x)\to-\infty\) as \(|x|\to\infty\). Because the exponential function is strictly increasing, \(g(x)\leq e\), and \(g(x)>0\). Thus, the range is \((0, e]\). 3. The graph of \(f\) increases for \(x<0\) and decreases for \(x>0\). Since \(e^u\) preserves order, \(g\) has the same monotonicity: it increases on \((-\infty, 0]\) and decreases on \([0, \infty)\). 4. Equivalently, \(g'(x)=f'(x)e^{f(x)}\), and the exponential factor is always positive, so the sign of \(g'\) is the sign of \(f'\).

a) \(\mathbb{R}\) b) \((0, e]\) c) Increasing on \((-\infty, 0]\) and decreasing on \([0, \infty)\)

- Use point-slope form for each tangent line. - Use symmetry about the \(y\)-axis. - Find the triangle area from its base and height. - The \(x\)-intercepts of the tangent lines determine the base.

1. Let \(c=\phi(1)=\phi(-1)=\frac{1}{\sqrt{2\pi e}}\). Since \(\phi'(x)=-x\phi(x)\), the slopes at \(-1\) and \(1\) are \(c\) and \(-c\), respectively. 2. a) At \((-1, c)\), the tangent line is \(y-c=c(x+1)\), so \(y=c(x+2)\). At \((1, c)\), the tangent line is \(y-c=-c(x-1)\), so \(y=-c(x-2)\). 3. b) The lines meet the \(x\)-axis at \((-2, 0)\) and \((2, 0)\), so the base has length \(4\). The tangent lines intersect at \((0, 2c)\), so the height is \(2c\). Therefore, \(A=\frac12\cdot4\cdot2c=4c=\frac{4}{\sqrt{2\pi e}}\approx0.968\).

a) \(y=\frac{1}{\sqrt{2\pi e}}(x+2)\) and \(y=-\frac{1}{\sqrt{2\pi e}}(x-2)\) b) \(A=\frac{4}{\sqrt{2\pi e}}\approx0.968\) square units

- Tangent graphs have equal values and equal slopes at the same input. - Differentiate both functions. - Use the double-angle identity for sine. - Solve the derivative equation first, then use the value equation.

1. At a point of tangency, the function values and derivatives must be equal: \(f(x)=g_c(x)\) and \(f'(x)=g_c'(x)\). 2. The derivatives are \(f'(x)=2\cos x\) and \(g_c'(x)=-2\sin(2x)\). Set them equal and use \(\sin(2x)=2\sin x\cos x\): \(2\cos x=-4\sin x\cos x\), so \(2\cos x(1+2\sin x)=0\). 3. If \(\cos x=0\), then \(\sin x=1\) or \(-1\), and \(\cos(2x)=-1\). For \(\sin x=1\), the value equation gives \(2=-1+c\), so \(c=3\). For \(\sin x=-1\), it gives \(-2=-1+c\), so \(c=-1\). 4. If \(\sin x=-\frac{1}{2}\), then \(\cos(2x)=\frac{1}{2}\). The value equation gives \(-1=\frac{1}{2}+c\), so \(c=-\frac{3}{2}\).

\(c\in\left\{-\frac{3}{2},-1,3\right\}\)

- Tangency requires equal function values and equal derivatives. - Use the chain rule to differentiate \(\sin^2x\). - Factor the derivative equation. - Use \(\sin^2x+\cos^2x=1\) in the remaining case.

1. At a point of tangency, \(f(x)=h_k(x)\) and \(f'(x)=h_k'(x)\). 2. Differentiate: \(f'(x)=2\sin x\cos x\) and \(h_k'(x)=\sin x\). Set the slopes equal: \(\sin x(2\cos x-1)=0\). 3. If \(\sin x=0\), then \(\cos x=1\) or \(-1\). When \(\cos x=1\), the value equation gives \(0=k-1\), so \(k=1\). When \(\cos x=-1\), it gives \(0=k+1\), so \(k=-1\). 4. If \(\cos x=\frac{1}{2}\), then \(\sin^2x=1-\frac{1}{4}=\frac{3}{4}\). The value equation gives \(\frac{3}{4}=k-\frac{1}{2}\), so \(k=\frac{5}{4}\).

\(k\in\left\{-1,1,\frac{5}{4}\right\}\)

- Identify the inner and outer functions in each composition. - Use the chain rule: derivative of the outer function times derivative of the inner function. - Recall how the parameters in \(a\sin(bx+c)+d\) affect amplitude, period, and shifts. - Use the minimum and maximum values of sine to determine the range.

1. The compositions are \(h_1(x)=\sin(mx+c)\) and \(h_2(x)=m\sin(x)+c\). 2. By the chain rule, \(h_1'(x)=m\cos(mx+c)\) and \(h_2'(x)=m\cos(x)\). 3. The amplitude of \(h_1\) remains \(1\). Its period is \(\frac{2\pi}{m}\). Since \(mx+c=m\left(x+\frac{c}{m}\right)\), the graph is shifted horizontally by \(-\frac{c}{m}\): left when \(c>0\) and right when \(c<0\). 4. Since \(-1\le \sin(x)\le 1\) and \(m>0\), the range of \(h_2\) is \([c-m, c+m]\). Set the endpoints equal to the required range: \(c-m=-2\), \(c+m=4\). Adding gives \(2c=2\), so \(c=1\). Then \(m=3\).

1. \(h_1(x)=\sin(mx+c)\); \(h_2(x)=m\sin(x)+c\) 2. \(h_1'(x)=m\cos(mx+c)\); \(h_2'(x)=m\cos(x)\) 3. The period is \(\frac{2\pi}{m}\), the amplitude is \(1\), and the horizontal shift is \(-\frac{c}{m}\). 4. \(m=3\) and \(c=1\)

- Use the identity \(\sin(-z)=-\sin(z)\). - Remember the inner derivative when differentiating \(f(-x)\). - Recall the symmetry identity for cosine. - Consider how the product of two even functions behaves.

1. Since \(f\) is odd, \(f(-x)=-f(x)\). Therefore, \(h(-x)=\sin(f(-x))=\sin(-f(x))=-\sin(f(x))=-h(x)\). Thus, \(h\) is odd. 2. By the chain rule, \(h'(x)=\cos(f(x))f'(x)\). 3. Differentiate \(f(-x)=-f(x)\) with respect to \(x\): \(-f'(-x)=-f'(x)\). Hence, \(f'(-x)=f'(x)\), so \(f'\) is even. 4. Now, \(h'(-x)=\cos(f(-x))f'(-x)\) \(=\cos(-f(x))f'(x)\) \(=\cos(f(x))f'(x)=h'(x)\). Therefore, \(h'\) is even.

a) \(h(-x)=\sin(f(-x))=\sin(-f(x))=-h(x)\), so \(h\) is odd. b) \(h'(x)=\cos(f(x))f'(x)\) c) Differentiating \(f(-x)=-f(x)\) gives \(-f'(-x)=-f'(x)\), so \(f'(-x)=f'(x)\). Thus, \(f'\) is even, and \(h'(-x)=\cos(-f(x))f'(x)=h'(x)\), so \(h'\) is even.

- First treat \(g(h(x))\) as one inner function. - Apply the chain rule again when differentiating that inner function. - Evaluate the table from the inside out: begin with \(h(0)\), then evaluate \(g\), and finally evaluate the needed derivative of \(f\). - Distinguish carefully between function values and derivative values.

1. Applying the chain rule to \(k(x)=f(V(x))\) gives \(k'(x)=f'(V(x))V'(x)\). 2. Since \(V(x)=g(h(x))\), another application of the chain rule gives \(V'(x)=g'(h(x))h'(x)\). Therefore, \(k'(x)=f'(g(h(x)))g'(h(x))h'(x)\). 3. At \(x=0\), the table gives \(h(0)=1\) and \(h'(0)=5\). 4. Then \(g(h(0))=g(1)=0\), and the table gives \(g'(1)=3\) and \(f'(0)=1\). 5. Thus, \(k'(0)=f'(0)g'(1)h'(0)=1\cdot 3\cdot 5=15\).

a) \(k'(x)=f'(g(h(x)))g'(h(x))h'(x)\) b) \(k'(0)=15\)

- Use the y-intercept condition before differentiating. - Relate the stated angle with the y-axis to a line's angle of inclination. - Evaluate the derivative at the y-intercept and compare that slope with the possible line directions. - Remember to enforce the restriction on \(k\).

1. The y-intercept gives \(f_{a,k}(0)=a=3\). 2. Differentiate using the product and chain rules: \(f_{a,k}'(x)=(1+kx+ak)e^{kx}\). 3. At \(x=0\), the tangent slope is \(1+3k\). 4. A line making a \(45^\circ\) angle with the y-axis also makes a \(45^\circ\) or \(135^\circ\) angle of inclination with the x-axis, so its slope is \(1\) or \(-1\). 5. The equation \(1+3k=1\) gives \(k=0\), which is excluded. The equation \(1+3k=-1\) gives \(k=-\frac{2}{3}\). 6. Therefore, \(a=3\) and \(k=-\frac{2}{3}\).

\(a=3\) and \(k=-\frac{2}{3}\)

- Apply both the quotient rule and the chain rule. - Factor \(u(x)\) from the numerator of the derivative. - Substitute \(u(x_0)=0\) only after differentiating. - Use \(v(x_0)\ne 0\) to confirm the derivative is defined.

1. Apply the quotient rule and chain rule: \(h'(x)=\frac{2u(x)u'(x)v(x)-[u(x)]^2v'(x)}{[v(x)]^2}\). 2. Factor \(u(x)\) from the numerator: \(h'(x)=\frac{u(x)[2u'(x)v(x)-u(x)v'(x)]}{[v(x)]^2}\). 3. Evaluate at \(x=x_0\). Since \(u(x_0)=0\), \(h'(x_0)=\frac{0[2u'(x_0)v(x_0)-0\cdot v'(x_0)]}{[v(x_0)]^2}=0\). Because \(v(x_0)\ne 0\), the denominator is defined and nonzero. Therefore, the graph of \(h\) has a horizontal tangent at \(x=x_0\).

\(h'(x)=\frac{u(x)[2u'(x)v(x)-u(x)v'(x)]}{[v(x)]^2}\). Since \(u(x_0)=0\) and \(v(x_0)\ne 0\), \(h'(x_0)=0\), so the tangent is horizontal.

- Substitute the given point into the function to form one equation. - Use the derivative and the given slope to form a second equation. - Solve the two equations for \(a\) and \(c\). - The expression \(x-c\) controls the horizontal shift, and \(a\) controls vertical scaling.

1. The point condition gives \(a\sqrt{5-c}=4\). 2. Since \(h'(x)=\frac{a}{2\sqrt{x-c}}\), the slope condition gives \(\frac{a}{2\sqrt{5-c}}=1\), or \(a=2\sqrt{5-c}\). 3. Substitute into the point condition: \(2(5-c)=4\). Thus, \(c=3\), and then \(a=2\sqrt{2}\). 4. The graph is shifted \(3\) units right and vertically stretched by a factor of \(2\sqrt{2}\).

a) \(a=2\sqrt{2}\); \(c=3\) b) Shift right \(3\) units and vertically stretch by a factor of \(2\sqrt{2}\).

- A normal slope is the negative reciprocal of the tangent slope when both are finite and nonzero. - Write a general normal equation at \(x_0\), and test whether \(M\) satisfies it. - Treat horizontal, vertical, and endpoint tangents as separate cases. - A tangent parallel to \(y=x\) has slope \(1\). - Check solutions in the original radical equation after squaring.

1. a) Require \(x(12-x)\ge0\), so \(D_g=[0,12]\). For \(0<x<12\), \(g'(x)=\frac{6-x}{\sqrt{12x-x^2}}\). 2. At \(x=2\), \(g(2)=2\sqrt{5}\), and the tangent slope is \(\frac{2}{\sqrt{5}}\). The normal slope is \(-\frac{\sqrt{5}}{2}\), so \(n:y=-\frac{\sqrt{5}}{2}x+3\sqrt{5}\). 3. b) For \(0<x_0<12\) and \(x_0\ne6\), the normal slope is \(m_n=\frac{\sqrt{12x_0-x_0^2}}{x_0-6}\). Substituting \(M=(6,0)\) into the normal equation gives \(0=m_n(6-x_0)+g(x_0)\), which is true. 4. At \(x_0=6\), the tangent is horizontal, so the normal is \(x=6\), which contains \(M\). At the endpoints \((0,0)\) and \((12,0)\), the tangents are vertical, so the normal is the x-axis, which also contains \(M\). 5. c) Parallelism to \(y=x\) requires \(g'(x)=1\). Thus, \(6-x=\sqrt{12x-x^2}\). Squaring gives \(x^2-12x+18=0\), with candidates \(x=6\pm3\sqrt{2}\). 6. The original unsquared equation requires \(x<6\), so \(x=6-3\sqrt{2}\). Then \(g(x)=3\sqrt{2}\), giving \(P=(6-3\sqrt{2},3\sqrt{2})\).

a) \(D_g=[0,12]\); \(n:y=-\frac{\sqrt{5}}{2}x+3\sqrt{5}\) b) Every normal contains \(M=(6,0)\); the interior, center, and endpoint cases all satisfy the claim. c) \(P=(6-3\sqrt{2},3\sqrt{2})\)

- Use the point-slope form of a tangent line at \(x_0\). - Differentiate the square-root function with the chain rule. - Set \(x=0\) to find a y-intercept. - Two points determine a unique line.

1. Differentiate: \(f'(x)=\frac{4}{\sqrt{8x}}\). 2. a) The tangent line is \(t(x)=\frac{4}{\sqrt{8x_0}}(x-x_0)+\sqrt{8x_0}\), which simplifies to \(t(x)=\frac{4}{\sqrt{8x_0}}x+\frac{1}{2}\sqrt{8x_0}\). 3. b) Setting \(x=0\) gives \(t(0)=\frac{1}{2}\sqrt{8x_0}=\frac{1}{2}f(x_0)\), so the y-intercept is \(Q\). 4. c) Project \(P\) horizontally to the y-axis to locate \((0,f(x_0))\). Construct the midpoint of the segment from the origin to that point; this midpoint is \(Q\). Draw the line through \(P\) and \(Q\).

a) \(t(x)=\frac{4}{\sqrt{8x_0}}x+\frac{1}{2}\sqrt{8x_0}\) b) \(t(0)=\frac{1}{2}f(x_0)\) c) Locate the point at half the height of \(P\) on the y-axis, then draw the line through that point and \(P\).

- Rewrite a variable base and exponent using the exponential function. - Differentiate the exponent using the product and chain rules. - Apply the chain rule to the outer exponential function.

1. Use \(a^b=e^{b\ln a}\): \(f(x)=e^{x\ln(\ln x)}\). Thus, \(g(x)=x\ln(\ln x)\). 2. Differentiate \(g\) using the product and chain rules: \(g'(x)=\ln(\ln x)+\frac{1}{\ln x}\). 3. Differentiate \(f(x)=e^{g(x)}\): \(f'(x)=e^{g(x)}g'(x)\). Therefore, \(f'(x)=(\ln x)^x\left(\ln(\ln x)+\frac{1}{\ln x}\right)\).

a) \(f(x)=e^{x\ln(\ln x)}\) b) \(f'(x)=(\ln x)^x\left(\ln(\ln x)+\frac{1}{\ln x}\right)\)

- Apply the chain rule to the logarithm of a function. - Think about how a strictly increasing function affects the order of two input values. - Determine whether the outer function reverses or preserves inequalities. - Relate the sign of the outer derivative to the derivative of the composition.

1. By the chain rule, \(h'(x)=\frac{f'(x)}{f(x)}\). If \(f'(x_0)=0\), then \(h'(x_0)=\frac{0}{f(x_0)}=0\) because \(f(x_0)>0\). Thus, \(x_0\) is a critical point of \(h\). 2. The function \(T\) must be strictly increasing on the range of values under consideration. A strictly increasing transformation preserves the order of function values, so it preserves both the location and type of local extrema. The function \(T(y)=y^2\) is strictly increasing for \(y>0\), so it preserves local maxima of a positive function. The function \(T(y)=\ln(y)\) is strictly increasing on \((0,\infty)\), so it also preserves local maxima of a positive function.

1) Since \(h'(x)=\frac{f'(x)}{f(x)}\) and \(f(x)>0\), \(f'(x_0)=0\) implies \(h'(x_0)=0\). 2) \(T\) must be strictly increasing on the relevant range. Both \(T(y)=y^2\) for \(y>0\) and \(T(y)=\ln(y)\) for \(y>0\) have this property, so they preserve local maxima of positive functions.

- First determine the domain of each square-root function. - Use the chain rule on the square root. - The sign of the derivative is controlled by its numerator. - Eliminate \(k\) from the maximum coordinates.

1. The radicand is nonnegative on \([0, 4k]\). Every graph contains \((0, 0)\). If a point with \(x\neq0\) were common to graphs with distinct parameters \(k\) and \(m\), then \(kx-\frac{x^2}{4}=mx-\frac{x^2}{4}\), which would imply \(k=m\). Thus the origin is the only common point. 2. For interior points of the domain, the chain rule gives \(f_k'(x)=\frac{k-x/2}{2\sqrt{kx-x^2/4}}\). The derivative is zero at \(x=2k\), positive before this point, and negative after it, so the graph has a local maximum there. 3. Its y-coordinate is \(f_k(2k)=\sqrt{2k^2-k^2}=k\). Therefore, the local maximum is \((2k, k)\). 4. Since \(x=2k\) and \(y=k\), all maxima lie on \(y=\frac{1}{2}x\).

a) \((0, 0)\) b) \((2k, k)\) c) \(y=\frac{1}{2}x\)

- Apply the chain rule to the composition. - A product is zero when at least one factor is zero. - Identify horizontal tangents on both graphs. - Find where \(v\) takes the input values at which \(u\) has horizontal tangents, using its equation for exact values.

1. The graph gives \(v(1)=1\) and \(u(1)=-1\), so \(g(1)=-1\). 2. By the chain rule, \(g'(x)=u'(v(x))v'(x)\). Since \(u\) has a horizontal tangent at input \(1\), \(u'(1)=0\). Therefore, \(g'(1)=0\). 3. The product \(u'(v(x))v'(x)\) is zero when either factor is zero. The graph of \(v\) has a horizontal tangent at \(x=0\), giving one zero. 4. The graph of \(u\) has horizontal tangents at inputs \(-1\) and \(1\). Solve \(v(x)=1\) and \(v(x)=-1\) using \(v(x)=2-x^2\). These give \(x=\pm1\) and \(x=\pm\sqrt{3}\), respectively. 5. Thus the five zeros are \(-\sqrt{3}\), \(-1\), \(0\), \(1\), and \(\sqrt{3}\).

a) \(g(1)=-1\) and \(g'(1)=0\) b) \(x\in\{-\sqrt{3}, -1, 0, 1, \sqrt{3}\}\)