Free math worksheets

- The second derivative is the derivative of the first derivative. - Differentiate each term of the given polynomial once. - Constants disappear when differentiated.

1. Differentiate the first derivative once more: \(f^{\prime\prime}(x)=12x\).

\(f^{\prime\prime}(x)=12x\)

- Interpret \(f^{\prime\prime}\) as the derivative of \(f^{\prime}\). - For a straight-line graph, the derivative is its slope. - Use two grid points to determine rise over run.

1. The value \(f^{\prime\prime}(0)\) is the slope of the graph of \(f^{\prime}\) at \(x=0\). 2. The displayed graph is a line that rises \(2\) units for each \(1\)-unit run, so its slope is \(2\). 3. Therefore, \(f^{\prime\prime}(0)=2\).

\(2\)

- Apply the power rule to each term. - What does the double-prime notation mean? - To evaluate a derivative at a point, first find its formula and then substitute the input.

1. Differentiate term by term: \(\frac{dy}{dx}=x^3-\frac{5}{2}x^2+2x\). 2. First, \(f'(x)=\frac{1}{4}x^4-\frac{3}{2}x^2+7\). Differentiate again: \(f''(x)=x^3-3x\). 3. First, \(f'(x)=\frac{1}{2}x^2-2x+5\), so \(f''(x)=x-2\). Therefore, \(f''(2)=0\).

1) \(\frac{dy}{dx}=x^3-\frac{5}{2}x^2+2x\) 2) \(f''(x)=x^3-3x\) 3) \(f''(2)=0\)

- Apply the power rule repeatedly. - Each differentiation lowers the degree of a nonconstant polynomial by \(1\). - Constant terms disappear when differentiated. - Use each derivative as the starting point for the next one.

1. First derivative: \(f'(x)=x^3-3x^2+10x\). 2. Second derivative: \(f''(x)=3x^2-6x+10\). 3. Third derivative: \(f'''(x)=6x-6\). 4. Fourth derivative: \(f^{(4)}(x)=6\).

\(f'(x)=x^3-3x^2+10x\) \(f''(x)=3x^2-6x+10\) \(f'''(x)=6x-6\) \(f^{(4)}(x)=6\)

- Find the third derivative by differentiating three times in succession. - The notation \(g'''(2)\) means to evaluate the third derivative at \(x=2\). - Apply exponents before multiplication when evaluating.

1. Differentiate successively: \(g'(x)=-0.5x^4+6x^2-1\), \(g''(x)=-2x^3+12x\), and \(g'''(x)=-6x^2+12\). 2. Evaluate: \(g'''(2)=-6(2)^2+12=-12\).

a) \(g'''(x)=-6x^2+12\) b) \(g'''(2)=-12\)

- Each application of the power rule decreases a positive integer exponent by \(1\). - A constant term has derivative \(0\). - Keep constant coefficients when differentiating powers. - Obtain each higher derivative by differentiating the preceding derivative.

1. Differentiate once: \(f'(x)=\frac{1}{2}x^2-4x+5\). 2. Differentiate again: \(f''(x)=x-4\). 3. The third derivative is \(f'''(x)=1\). 4. The fourth derivative is \(f^{(4)}(x)=0\).

\(f'(x)=\frac{1}{2}x^2-4x+5\) \(f''(x)=x-4\) \(f'''(x)=1\) \(f^{(4)}(x)=0\)

- List derivatives of cosine until the pattern repeats. - Keep the constant factor \(5\). - Divide \(51\) by the cycle length and use the remainder. - Recall the exact value of \(\sin\pi\).

1. The derivatives of cosine repeat in a cycle of length \(4\): \(\cos x,-\sin x,-\cos x,\sin x,\cos x\). 2. The constant factor \(5\) remains unchanged. Since \(51=12\cdot4+3\), the 51st derivative matches the third derivative in the cycle: \(g^{(51)}(x)=5\sin x\). 3. Evaluate at \(x=\pi\): \(g^{(51)}(\pi)=5\sin\pi=0\).

a) \(g^{(51)}(x)=5\sin x\) b) \(g^{(51)}(\pi)=0\)

- Apply the power rule one derivative at a time. - Constant terms have derivative \(0\). - Keep constant coefficients when differentiating. - What is the derivative of a constant function?

1. \(f'(x)=\frac{1}{6}x^3-x+3\). 2. \(f''(x)=\frac{1}{2}x^2-1\). 3. \(f'''(x)=x\). 4. \(f^{(4)}(x)=1\). 5. \(f^{(5)}(x)=0\).

\(f'(x)=\frac{1}{6}x^3-x+3\) \(f''(x)=\frac{1}{2}x^2-1\) \(f'''(x)=x\) \(f^{(4)}(x)=1\) \(f^{(5)}(x)=0\)

- Apply the power rule to each term. - Constant terms disappear when differentiated. - At which derivative order does the variable disappear completely? - Track how the degree changes after each derivative.

1. Differentiate successively: \(f'(x)=4x^3+15x^2-2\), \(f''(x)=12x^2+30x\), \(f'''(x)=24x+30\), and \(f^{(4)}(x)=24\). 2. The fourth derivative is the constant function \(24\). 3. Since the derivative of a constant is \(0\), \(f^{(5)}(x)=0\).

1) \(f'(x)=4x^3+15x^2-2\), \(f''(x)=12x^2+30x\), \(f'''(x)=24x+30\), and \(f^{(4)}(x)=24\) 2) \(f^{(4)}(x)=24\) 3) \(f^{(5)}(x)=0\)

- Apply the power rule to every term. - Constant terms disappear when differentiated. - Multiply each coefficient by the exponent before reducing the exponent by \(1\). - Use each derivative as the starting point for the next derivative.

1. Differentiate once: \(f'(x)=10x^4-6x^3+16x-5\). 2. Differentiate again: \(f''(x)=40x^3-18x^2+16\). 3. The third derivative is \(f'''(x)=120x^2-36x\).

\(f'(x)=10x^4-6x^3+16x-5\) \(f''(x)=40x^3-18x^2+16\) \(f'''(x)=120x^2-36x\)

- Treat fractions as constant coefficients. - The symbol \(\pi\) represents a constant. - Simplify fractions after each differentiation step. - Check every term after each application of the power rule.

1. The first derivative is \(h'(x)=\frac{1}{12}x^4-\frac{1}{3}x^3+2x^2\). 2. The second derivative is \(h''(x)=\frac{1}{3}x^3-x^2+4x\). 3. The third derivative is \(h'''(x)=x^2-2x+4\).

\(h'(x)=\frac{1}{12}x^4-\frac{1}{3}x^3+2x^2\) \(h''(x)=\frac{1}{3}x^3-x^2+4x\) \(h'''(x)=x^2-2x+4\)

- Compute the first few derivatives. - Identify the factor introduced by each use of the chain rule. - Track the alternating sign. - Express repeated multiplication with an exponent.

1. The first derivatives are \(f'(x)=-2e^{5-2x}\) and \(f''(x)=(-2)^2e^{5-2x}\). 2. Each differentiation multiplies the expression by the derivative of the exponent, which is \(-2\). Therefore, after \(n\) derivatives, \(f^{(n)}(x)=(-2)^ne^{5-2x}\).

\(f^{(n)}(x)=(-2)^ne^{5-2x}\)

- Zeros of the first derivative occur at horizontal tangent lines. - Zeros of the second derivative can occur at inflection points. - Analyze each continuous branch separately.

1. On each interval of its domain, the graph is strictly decreasing and has no horizontal tangent lines. Therefore, \(f^{\prime}\) has \(0\) zeros. 2. Each branch keeps the same concavity and has no inflection point. The change across the vertical asymptote does not count because the function is undefined there. Therefore, \(f^{\prime\prime}\) has \(0\) zeros.

\(f^{\prime}\) has \(0\) zeros, and \(f^{\prime\prime}\) has \(0\) zeros.

- Determine how the constant factor changes with each derivative. - Treat the function itself as the zeroth derivative.

1. Base case: For \(n=0\), \(f^{(0)}(x)=f(x)=e^{2x}=2^0e^{2x}\). 2. Inductive hypothesis: Assume \(f^{(k)}(x)=2^ke^{2x}\) for an arbitrary integer \(k\ge0\). 3. Differentiate: \(f^{(k+1)}(x)=2^k\cdot2e^{2x}=2^{k+1}e^{2x}\). 4. Therefore, the formula holds for every integer \(n\ge0\).

For every integer \(n\ge0\), \(f^{(n)}(x)=2^ne^{2x}\).

- Differentiate repeatedly until the original function returns. - Count the number of derivatives in one full cycle. - Divide \(102\) by the cycle length. - Use the remainder to identify the matching derivative.

1. Differentiate repeatedly: \(f'(x)=\cos x\), \(f''(x)=-\sin x\), \(f'''(x)=-\cos x\), and \(f^{(4)}(x)=\sin x\). 2. The fourth derivative equals the original function, so the same four derivatives repeat. The cycle length is \(4\). 3. Since \(102=25\cdot 4+2\), the 102nd derivative matches the second derivative. Therefore, \(f^{(102)}(x)=-\sin x\).

1. \(f'(x)=\cos x\), \(f''(x)=-\sin x\), \(f'''(x)=-\cos x\), \(f^{(4)}(x)=\sin x\) 2. Cycle length: \(4\) 3. \(f^{(102)}(x)=-\sin x\)

- Write the first few derivatives of cosine. - Use the four-step derivative cycle. - Reduce \(55\) modulo \(4\). - Compare odd- and even-order derivatives at \(x=0\).

1. The first two derivatives are \(g'(x)=-\sin x\) and \(g''(x)=-\cos x=-g(x)\). Therefore, the smallest positive integer is \(n=2\). 2. The derivatives repeat every four steps. Since \(55=13\cdot 4+3\), \(g^{(55)}\) matches the third derivative. Thus, \(g^{(55)}(x)=\sin x\). 3. Odd-order derivatives are \(\pm\sin x\), which equal \(0\) at \(x=0\). Even-order derivatives are \(\pm\cos x\), which equal \(\pm1\) at \(x=0\). Therefore, \(n\in\{1,3,5,7,9\}\).

1. \(n=2\) 2. \(g^{(55)}(x)=\sin x\) 3. \(n\in\{1,3,5,7,9\}\)

- View the function as a product of two factors. - Apply the product rule to find the first derivative. - Treat \(k\) as a constant when differentiating. - Differentiate the first derivative again to obtain the second derivative. - Group the sine terms and the cosine terms in your final expression.

1. Let \(u(x) = kx^2 - 1\) and \(v(x) = \sin(x)\). Treat \(k\) as a constant, so \(u'(x) = 2kx\) and \(v'(x) = \cos(x)\). 2. Apply the product rule: \(f_k'(x) = 2kx\sin(x) + (kx^2 - 1)\cos(x)\). 3. Differentiate both terms again: \(\frac{d}{dx}[2kx\sin(x)] = 2k\sin(x) + 2kx\cos(x)\), \(\frac{d}{dx}[(kx^2 - 1)\cos(x)] = 2kx\cos(x) - (kx^2 - 1)\sin(x)\). 4. Combine like trigonometric terms: \(f_k''(x) = (2k - kx^2 + 1)\sin(x) + 4kx\cos(x)\).

\(f_k'(x) = 2kx\sin(x) + (kx^2 - 1)\cos(x)\) \(f_k''(x) = (2k - kx^2 + 1)\sin(x) + 4kx\cos(x)\)

- First check whether the two terms in the original function can be combined. - Use the product rule after simplifying. - Watch the signs when differentiating sine and cosine. - To find the second derivative, differentiate each term of the first derivative.

1. Combine like terms: \(g_a(t) = 3at^2\cos(t)\). 2. Let \(u(t) = 3at^2\) and \(v(t) = \cos(t)\). Then \(u'(t) = 6at\) and \(v'(t) = -\sin(t)\). 3. Apply the product rule: \(g_a'(t) = 6at\cos(t) - 3at^2\sin(t)\). 4. Differentiate both terms again: \(g_a''(t) = 6a\cos(t) - 6at\sin(t) - 6at\sin(t) - 3at^2\cos(t)\). 5. Combine like terms: \(g_a''(t) = (6a - 3at^2)\cos(t) - 12at\sin(t)\).

\(g_a'(t) = 6at\cos(t) - 3at^2\sin(t)\) \(g_a''(t) = (6a - 3at^2)\cos(t) - 12at\sin(t)\)

- Which rule applies to a product of two functions? - Remember that the derivative of \(e^x\) is \(e^x\). - When finding the second derivative, apply the product rule again to every product term. - How does the order of a derivative of \(g\) change when you differentiate it once more?

1. For a), apply the product rule: \(f'(x) = e^xg(x) + e^xg'(x) = e^x(g(x) + g'(x))\). 2. Differentiate again: \(f''(x) = e^x(g(x) + g'(x)) + e^x(g'(x) + g''(x))\) \(= e^x(g(x) + 2g'(x) + g''(x))\). 3. For b), apply the product rule: \(f'(x) = 2xg'(x) + (x^2 + 1)g''(x)\). 4. Differentiate both terms: \(\frac{d}{dx}[2xg'(x)] = 2g'(x) + 2xg''(x)\), \(\frac{d}{dx}[(x^2 + 1)g''(x)] = 2xg''(x) + (x^2 + 1)g'''(x)\). Therefore, \(f''(x) = 2g'(x) + 4xg''(x) + (x^2 + 1)g'''(x)\).

a) \(f'(x) = e^x(g(x) + g'(x))\) and \(f''(x) = e^x(g(x) + 2g'(x) + g''(x))\) b) \(f'(x) = 2xg'(x) + (x^2 + 1)g''(x)\) and \(f''(x) = 2g'(x) + 4xg''(x) + (x^2 + 1)g'''(x)\)

- First find every required derivative of \(u\) and \(v\). - When differentiating \(f'(x)\), keep track of the minus sign and use parentheses. - In part c), substitute the derivatives from part a) directly into the given formula.

1. The required derivatives are \(u'(x) = 3x^2\), \(u''(x) = 6x\), \(v'(x) = -\sin(x)\), and \(v''(x) = -\cos(x)\). 2. First differentiate \(f\): \(f'(x) = 3x^2\cos(x) - x^3\sin(x)\). 3. Differentiate again: \(f''(x) = 6x\cos(x) - 3x^2\sin(x) - [3x^2\sin(x) + x^3\cos(x)]\) \(= 6x\cos(x) - 6x^2\sin(x) - x^3\cos(x)\). 4. Substituting into the formula gives \((6x)\cos(x) + 2(3x^2)(-\sin(x)) + x^3(-\cos(x))\) \(= 6x\cos(x) - 6x^2\sin(x) - x^3\cos(x)\), which matches the result from part b).

a) \(u'(x) = 3x^2\), \(u''(x) = 6x\), \(v'(x) = -\sin(x)\), and \(v''(x) = -\cos(x)\) b) \(f''(x) = 6x\cos(x) - 6x^2\sin(x) - x^3\cos(x)\) c) Substitution into the given formula produces the same expression as in part b).

- Apply the product rule to the first few derivatives and simplify each result. - Determine how many derivatives it takes for cosine to return to itself. - Look for a relationship between the derivative order and the coefficient of the trigonometric term. - Compare the term containing \(x\) with the corresponding derivative of cosine.

1. Apply the product rule repeatedly: \(f'(x) = \cos(x) - x\sin(x)\), \(f''(x) = -2\sin(x) - x\cos(x)\), \(f'''(x) = -3\cos(x) + x\sin(x)\). 2. The derivatives of cosine repeat every four derivatives: \(\cos(x), -\sin(x), -\cos(x), \sin(x), \cos(x)\). Since \(40\) is divisible by \(4\), \(c^{(40)}(x) = \cos(x)\). 3. Because all derivatives of \(x\) after the first are zero, the generalized product rule reduces to \(f^{(n)}(x) = x\,c^{(n)}(x) + n c^{(n-1)}(x)\). For \(n=40\), \(c^{(40)}(x)=\cos(x)\) and \(c^{(39)}(x)=\sin(x)\). Therefore, \(f^{(40)}(x) = x\cos(x) + 40\sin(x)\).

1. \(f'(x) = \cos(x) - x\sin(x)\); \(f''(x) = -2\sin(x) - x\cos(x)\); \(f'''(x) = -3\cos(x) + x\sin(x)\) 2. \(c^{(40)}(x) = \cos(x)\) because the derivative cycle has period \(4\). 3. \(f^{(40)}(x) = x\cos(x) + 40\sin(x)\)

- Identify the inner and outer functions. - Include the derivative of the inner linear function. - Pay close attention to the negative sign. - Notice how the exponent changes with each derivative.

1. Apply the chain rule: \(f'(x)=6(4-5x)^5\cdot(-5)=-30(4-5x)^5\). 2. Differentiate again using the chain rule: \(f''(x)=-30\cdot 5(4-5x)^4\cdot(-5)\) \(=750(4-5x)^4\).

\(f'(x)=-30(4-5x)^5\) \(f''(x)=750(4-5x)^4\)

- Use the chain rule for the original composite function. - The first derivative is a product of two functions of \(x\). - Apply both the product rule and the chain rule when differentiating again. - Recall the derivatives of sine and cosine.

1. Apply the chain rule: \(f'(x)=2\cos(x^2+1)(2x)=4x\cos(x^2+1)\). 2. Differentiate the product \(4x\cos(x^2+1)\): \(f''(x)=4\cos(x^2+1)+4x[-\sin(x^2+1)(2x)]\). Therefore, \(f''(x)=4\cos(x^2+1)-8x^2\sin(x^2+1)\).

\(f'(x)=4x\cos(x^2+1)\) \(f''(x)=4\cos(x^2+1)-8x^2\sin(x^2+1)\)

- Apply the chain rule one derivative at a time. - Compute the first four derivatives and track both the sign and the trigonometric function. - The sine-cosine derivative cycle has length \(4\). - Check whether the requested derivative order is a multiple of the cycle length.

1. Differentiate using the chain rule: \(f_a'(x)=a\cos(ax+1)\). 2. Differentiate again: \(f_a''(x)=-a^2\sin(ax+1)\). 3. The derivatives repeat their trigonometric pattern every four derivatives: \(f_a^{(3)}(x)=-a^3\cos(ax+1)\), \(f_a^{(4)}(x)=a^4\sin(ax+1)\). Since \(40\) is divisible by \(4\), \(f_a^{(40)}(x)=a^{40}\sin(ax+1)\).

\(f_a''(x)=-a^2\sin(ax+1)\) \(f_a^{(40)}(x)=a^{40}\sin(ax+1)\)

- Use the chain rule for an exponential function with \(2x\) in the exponent. - Write the tangent equation at \(x=0\). - Negative slope gives a sign condition on \(k\). - Reverse an inequality when multiplying by a negative quantity. - Set the tangent equation equal to \(0\) to find its x-intercept.

1. Differentiate twice: \(f_k'(x)=2ke^{2x}\) and \(f_k''(x)=4ke^{2x}\). Therefore, \(f_k''(0)=4k\). 2. At \(x=0\), the point is \((0, k-4)\) and the slope is \(2k\). The tangent line is \(y=2kx+k-4\). 3. Negative slope requires \(k<0\). 4. The x-intercept is \(x=\frac{4-k}{2k}\). Require \(\frac{4-k}{2k}<-1\). 5. Because \(2k<0\), multiplying reverses the inequality: \(4-k>-2k\), so \(k>-4\). 6. Combining the conditions gives \(-4<k<0\).

a) \(f_k''(0)=4k\) b) \(-4<k<0\)

- Identify where the sum, product, and chain rules are needed. - Include the derivative of each exponent. - Factor out \(e^x\) after using the product rule. - Treat each parameter as a constant.

1. For part a, apply the chain rule to each exponential term: \(f_k'(x)=2ke^{2x}-ke^{-kx}\). 2. Differentiate again: \(f_k''(x)=4ke^{2x}+k^2e^{-kx}\). 3. For part b, apply the product rule: \(h_a'(x)=2xe^x+(x^2-a)e^x=(x^2+2x-a)e^x\). 4. Differentiate again: \(h_a''(x)=(2x+2)e^x+(x^2+2x-a)e^x=(x^2+4x+2-a)e^x\).

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

- A horizontal tangent has slope \(0\). - Use both the product rule and the chain rule. - The exponential factor cannot be zero. - Factor out \(e^{tx}\) after differentiating.

1. Apply the product and chain rules: \(g_t'(x)=(t^2-tx-1)e^{tx}\). 2. Differentiate again: \(g_t''(x)=(t^3-t^2x-2t)e^{tx}\). 3. A horizontal tangent requires \(g_t'(x)=0\). 4. Since \(e^{tx}>0\), solve \(t^2-tx-1=0\). 5. Because \(t\neq0\), \(x=\frac{t^2-1}{t}=t-\frac{1}{t}\).

a) \(g_t'(x)=(t^2-tx-1)e^{tx}\), \(g_t''(x)=(t^3-t^2x-2t)e^{tx}\) b) \(x=t-\frac{1}{t}\)

- Substitute the point into the original function. - Use the product rule for a linear factor times \(e^x\). - Evaluate the second derivative at the specified x-value.

1. Substitute \((1, 5e)\): \((k+4)e=5e\), so \(k=1\). 2. Differentiate: \(f_k'(x)=(kx+k+4)e^x\). 3. Differentiate again: \(f_k''(x)=(kx+2k+4)e^x\). 4. Set \(f_k''(0)=2\): \(2k+4=2\), so \(k=-1\).

1. \(k=1\) 2. \(k=-1\)

- Substitute the point into the original function. - Apply the product rule and factor out the exponential term. - An equation of the form \(a^2=c\) may have two real solutions.

1. The point condition gives \(g_a(0)=-a^2=-9\), so \(a^2=9\) and \(a=\pm3\). 2. Differentiate: \(g_a'(x)=(x^2+2x-a^2)e^x\). 3. Differentiate again: \(g_a''(x)=(x^2+4x+2-a^2)e^x\). 4. The condition \(g_a''(0)=-7\) gives \(2-a^2=-7\), so \(a^2=9\) and \(a=\pm3\).

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

- Differentiate the two exponential terms separately. - Look for the pattern in repeated derivatives of \(e^{ax}\). - Determine the sign of \((-4)^{21}\). - Factor out the common constant at the end.

1. For any positive integer \(n\), \(\frac{d^n}{dx^n}e^{ax}=a^ne^{ax}\). Therefore, \(f^{(n)}(x)=4^ne^{4x}-(-4)^ne^{-4x}\). 2. Since \(21\) is odd, \((-4)^{21}=-4^{21}\). Thus, \(f^{(21)}(x)=4^{21}e^{4x}+4^{21}e^{-4x}\) \(=4^{21}(e^{4x}+e^{-4x})\).

\(f^{(21)}(x)=4^{21}(e^{4x}+e^{-4x})\)

- Would expanding the product make differentiation easier? - Set the second derivative equal to zero to find the requested values. - An equation involving a square can have two real solutions.

1. Expand first: \(h(t)=6t^2-t^4\). Then \(h'(t)=12t-4t^3\). 2. Differentiate again: \(h''(t)=12-12t^2\). 3. Solve \(12-12t^2=0\). This gives \(t^2=1\), so \(t=-1\) or \(t=1\).

1) \(h'(t)=12t-4t^3\) 2) \(h''(t)=12-12t^2\) 3) \(t=-1\) and \(t=1\)

- Rewrite the rational function as a constant plus a power of \(x-3\). - Apply the power rule twice. - Examine the end behavior of the second derivative. - Use the multiplicity of the denominator factor to determine the sign behavior near the vertical asymptote.

1. Rewrite the function as \(f(x)=5+\frac{17}{x-3}\). 2. Differentiate: \(f'(x)=-17(x-3)^{-2}=\frac{-17}{(x-3)^2}\). 3. Differentiate again: \(f''(x)=34(x-3)^{-3}=\frac{34}{(x-3)^3}\). 4. Since \(\lim_{x\to\pm\infty}\frac{34}{(x-3)^3}=0\), the x-axis, \(y=0\), is a horizontal asymptote. 5. The denominator is zero at \(x=3\), so \(x=3\) is a vertical asymptote. Because the denominator has odd multiplicity, \(f''\) changes sign across \(x=3\).

a) \(f''(x)=\frac{34}{(x-3)^3}\) b) \(\lim_{x\to\pm\infty}f''(x)=0\), so \(y=0\) is a horizontal asymptote. c) The vertical asymptote is \(x=3\), and \(f''\) changes sign across it.

- Rewrite the function using a negative exponent. - Apply the power rule and chain rule twice. - Examine the end behavior of the second derivative. - Use the even exponent in the denominator to determine the sign on both sides of the vertical asymptote.

1. Rewrite the function as \(g(x)=(x+2)^{-2}\). 2. Differentiate: \(g'(x)=-2(x+2)^{-3}=\frac{-2}{(x+2)^3}\). 3. Differentiate again: \(g''(x)=6(x+2)^{-4}=\frac{6}{(x+2)^4}\). 4. Since \(\lim_{x\to\infty}g''(x)=0\) and \(\lim_{x\to-\infty}g''(x)=0\), the x-axis, \(y=0\), is a horizontal asymptote. 5. The denominator is zero at \(x=-2\), so \(x=-2\) is a vertical asymptote. Because the denominator has even multiplicity and the numerator is positive, \(g''(x)>0\) on both sides of the asymptote. Thus, there is no sign change.

a) \(g''(x)=\frac{6}{(x+2)^4}\) b) \(\lim_{x\to\pm\infty}g''(x)=0\), so \(y=0\) is a horizontal asymptote. c) The vertical asymptote is \(x=-2\), and \(g''\) does not change sign across it.

- Rewrite the reciprocal with a negative exponent before differentiating. - Track how the exponent of \(x-1\) changes with each derivative. - For the sign question, compare the sign of an odd power of \(x-1\) on the two sides of \(1\).

1. Rewrite \(f(x)=(x-1)^{-1}\). Then \(f^{\prime}(x)=-(x-1)^{-2}=-\frac{1}{(x-1)^2}\). Differentiate again: \(f^{\prime\prime}(x)=2(x-1)^{-3}=\frac{2}{(x-1)^3}\). 2. The pole orders of \(f\), \(f^{\prime}\), and \(f^{\prime\prime}\) are \(1\), \(2\), and \(3\), respectively. Each differentiation multiplies by the current negative exponent and lowers that exponent by \(1\), so \(f^{(m)}(x)=\frac{(-1)^m m!}{(x-1)^{m+1}}\). Therefore, \(f^{(m)}\) has pole order \(m+1\) at \(x=1\). 3. Because \(f^{\prime\prime}(x)=\frac{2}{(x-1)^3}\), its denominator changes sign across \(x=1\). Thus \(\lim_{x\to1^-}f^{\prime\prime}(x)=-\infty\) and \(\lim_{x\to1^+}f^{\prime\prime}(x)=+\infty\), so the graph changes sign across the asymptote.

1. \(f^{\prime}(x)=-\frac{1}{(x-1)^2}\) and \(f^{\prime\prime}(x)=\frac{2}{(x-1)^3}\) 2. The pole orders are \(1\), \(2\), and \(3\). In general, \(f^{(m)}(x)=\frac{(-1)^m m!}{(x-1)^{m+1}}\), with pole order \(m+1\). 3. Yes. \(f^{\prime\prime}\) changes from negative to positive across \(x=1\).

- Exclude values that make the denominator zero. - Apply the quotient rule before differentiating again. - An odd power in the denominator produces opposite signs on the two sides of the vertical asymptote. - Find the x-intercept from the numerator, then use point-slope form.

1. The denominator is zero at \(x=-\frac{1}{2}\), so the domains of \(f\) and its derivatives exclude that value. 2. Apply the quotient rule: \(f'(x)=\frac{-3(4x+2)-4(6-3x)}{(4x+2)^2}=\frac{-30}{(4x+2)^2}\). 3. Differentiate again: \(f''(x)=\frac{240}{(4x+2)^3}\). Therefore, the domain is \(\mathbb{R}\setminus\left\{-\frac{1}{2}\right\}\), and \(x=-\frac{1}{2}\) is a vertical asymptote. Because the denominator has odd power \(3\), \(f''\) changes sign across the asymptote. 4. The x-intercept satisfies \(6-3x=0\), so it is \((2,0)\). The slope there is \(f'(2)=\frac{-30}{10^2}=-\frac{3}{10}\). 5. Using point-slope form, \(y=-\frac{3}{10}(x-2)\), or \(y=-\frac{3}{10}x+\frac{3}{5}\).

a) \(D_{f''}=\mathbb{R}\setminus\left\{-\frac{1}{2}\right\}\). The vertical asymptote is \(x=-\frac{1}{2}\), and \(f''\) changes sign across it. b) \(y=-\frac{3}{10}x+\frac{3}{5}\).

- Apply the product rule twice. - Set the function equal to its second derivative. - Divide by factors that are never zero.

1. Using the product rule, \(f_k'(x)=k(x^2+2x-1)e^x\) and \(f_k''(x)=k(x^2+4x+1)e^x\). 2. At an intersection, \(k(x^2-1)e^x=k(x^2+4x+1)e^x\). 3. Since \(k\ne0\) and \(e^x>0\), divide by \(ke^x\): \(x^2-1=x^2+4x+1\). 4. Solving gives \(x=-\frac{1}{2}\), which does not depend on \(k\).

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

- Apply the quotient rule to the original function. - Simplify the first derivative before differentiating again. - Use the chain rule when differentiating the squared denominator. - Factor before canceling common factors.

1. By the quotient rule, \(f'(x)=\frac{4(x^2+1)-4x(2x)}{(x^2+1)^2}\) \(=\frac{4-4x^2}{(x^2+1)^2}\). 2. Differentiate the first derivative using the quotient rule: \(f''(x)=\frac{-8x(x^2+1)^2-(4-4x^2)2(x^2+1)(2x)}{(x^2+1)^4}\). 3. Factor and cancel one factor of \(x^2+1\): \(f''(x)=\frac{-8x(x^2+1)-4x(4-4x^2)}{(x^2+1)^3}\). 4. Simplifying the numerator gives \(f''(x)=\frac{8x^3-24x}{(x^2+1)^3}\).

\(f'(x)=\frac{4-4x^2}{(x^2+1)^2}\) \(f''(x)=\frac{8x^3-24x}{(x^2+1)^3}\)

- Use the quotient rule for each derivative. - Simplify the first derivative before differentiating again. - Look for a common factor of \(x-1\) before expanding. - Apply the chain rule to the squared denominator.

1. By the quotient rule, \(g'(x)=\frac{(2x+2)(x-1)-(x^2+2x)}{(x-1)^2}\) \(=\frac{x^2-2x-2}{(x-1)^2}\). 2. Differentiate again: \(g''(x)=\frac{(2x-2)(x-1)^2-(x^2-2x-2)2(x-1)}{(x-1)^4}\). 3. Cancel one factor of \(x-1\): \(g''(x)=\frac{(2x-2)(x-1)-2(x^2-2x-2)}{(x-1)^3}\). 4. The numerator simplifies to \(6\), so \(g''(x)=\frac{6}{(x-1)^3}\).

\(g'(x)=\frac{x^2-2x-2}{(x-1)^2}\) \(g''(x)=\frac{6}{(x-1)^3}\)

- Decide whether the radical is in the numerator or denominator before choosing the sign of the exponent. - Apply the power rule one derivative at a time. - The denominator of a rational exponent determines the index of the radical. - Watch the signs when differentiating negative exponents.

1. The first line has the wrong sign on the exponent. Because the fourth root is in the denominator, \(f(x)=2x^{-\frac{1}{4}}\). 2. Differentiate correctly: \(f'(x)=2\left(-\frac{1}{4}\right)x^{-\frac{5}{4}}=-\frac{1}{2}x^{-\frac{5}{4}}\). 3. Differentiate again: \(f''(x)=-\frac{1}{2}\left(-\frac{5}{4}\right)x^{-\frac{9}{4}}=\frac{5}{8}x^{-\frac{9}{4}}=\frac{5}{8\sqrt[4]{x^9}}\). The student also converted \(x^{-\frac{7}{4}}\) incorrectly; it would equal \(\frac{1}{\sqrt[4]{x^7}}\), not \(\sqrt[7]{x^4}\).

The student used a positive exponent instead of a negative exponent when rewriting the reciprocal and later converted a negative rational exponent to radical form incorrectly. The correct result is \(f''(x)=\frac{5}{8}x^{-\frac{9}{4}}=\frac{5}{8\sqrt[4]{x^9}}\).

- First rewrite the radical and reciprocal as powers of \(x\). - Apply the power rule twice, one derivative at a time. - Be careful when subtracting \(1\) from fractional and negative exponents. - A negative rational exponent can be rewritten as a radical in the denominator.

1. Rewrite the function as \(f(x)=6x^{\frac{2}{3}}-4x^{-1}\). 2. Differentiate once: \(f'(x)=4x^{-\frac{1}{3}}+4x^{-2}\). 3. Differentiate again: \(f''(x)=-\frac{4}{3}x^{-\frac{4}{3}}-8x^{-3}\). 4. Rewrite without rational exponents: \(f''(x)=-\frac{4}{3\sqrt[3]{x^4}}-\frac{8}{x^3}\).

\(f''(x)=-\frac{4}{3\sqrt[3]{x^4}}-\frac{8}{x^3}\)

- Use the radicand condition for the original domain. - Apply the chain rule first and then differentiate again carefully. - Use the graph point and derivative value to write the tangent line. - Find the line's x- and y-intercepts to identify the triangle's legs. - The area of a right triangle is one-half the product of its leg lengths.

1. a) The radicand condition gives \(D_f=[-10,10]\). By the chain rule, \(f'(x)=-\frac{x}{\sqrt{100-x^2}}\), with \(D_{f'}=(-10,10)\). 2. Differentiate again: \(f''(x)=-\frac{100}{(100-x^2)^{\frac{3}{2}}}\), with \(D_{f''}=(-10,10)\). 3. b) At \(x=6\), \(f(6)=8\) and \(f'(6)=-\frac{3}{4}\). Therefore, \(t:y=-\frac{3}{4}x+\frac{25}{2}\). 4. c) The y-intercept is \(\left(0,\frac{25}{2}\right)\), and the x-intercept is \(\left(\frac{50}{3},0\right)\). 5. The triangle area is \(A=\frac{1}{2}\left(\frac{50}{3}\right)\left(\frac{25}{2}\right)=\frac{625}{6}\approx104.17\) square units.

a) \(f'(x)=-\frac{x}{\sqrt{100-x^2}}\); \(f''(x)=-\frac{100}{(100-x^2)^{\frac{3}{2}}}\); \(D_f=[-10,10]\); \(D_{f'}=D_{f''}=(-10,10)\) b) \(t:y=-\frac{3}{4}x+\frac{25}{2}\) c) \(A=\frac{625}{6}\approx104.17\) square units

- Use the chain rule for the square root. - Differentiate the first derivative carefully and simplify. - Evaluate the second derivative at \(x=0\). - Use the point-slope form of a tangent line.

1. Apply the chain rule: \(f_k'(x)=\frac{x}{\sqrt{x^2+k}}\). 2. Differentiate again: \(f_k''(x)=\frac{k}{(x^2+k)^{3/2}}\). 3. At \(x=0\), \(f_k''(0)=\frac{1}{\sqrt{k}}\). Set \(\frac{1}{\sqrt{k}}=\frac{1}{4}\), giving \(k=16\). 4. For \(k=12\), \(f_{12}(2)=4\) and \(f_{12}'(2)=\frac{1}{2}\). 5. The tangent line is \(y-4=\frac{1}{2}(x-2)\), or \(y=\frac{1}{2}x+3\).

a) \(f_k'(x)=\frac{x}{\sqrt{x^2+k}}\), \(f_k''(x)=\frac{k}{(x^2+k)^{3/2}}\) b) \(k=16\) c) \(y=\frac{1}{2}x+3\)

- The radicand is always positive. - Differentiate the first derivative to obtain the second derivative. - Simplify before substituting \(x=0\). - Apply the condition \(k>0\).

1. Since \(x^2+k^2>0\) for every real \(x\), the domain is \(\mathbb{R}\). 2. The first derivative is \(g_k'(x)=\frac{x}{k\sqrt{x^2+k^2}}\). 3. Differentiate again and simplify: \(g_k''(x)=\frac{k}{(x^2+k^2)^{3/2}}\). 4. At \(x=0\), \(g_k''(0)=\frac{k}{k^3}=\frac{1}{k^2}\). 5. Set \(\frac{1}{k^2}=4\). Since \(k>0\), \(k=\frac{1}{2}\).

a) Domain: \(\mathbb{R}\) b) \(g_k''(x)=\frac{k}{(x^2+k^2)^{3/2}}\) c) \(k=\frac{1}{2}\)

- Use the chain rule for the first derivative. - The first derivative is a quotient. - Track signs carefully when simplifying the numerator.

1. Apply the chain rule: \(f'(x)=\frac{2x}{x^2+4}\). 2. Differentiate using the quotient rule: \(f''(x)=\frac{2(x^2+4)-2x(2x)}{(x^2+4)^2}\). 3. Simplify: \(f''(x)=\frac{8-2x^2}{(x^2+4)^2}\).

\(f''(x)=\frac{8-2x^2}{(x^2+4)^2}\)

- Find the first derivative using the chain rule. - Write the first derivative as a quotient. - Apply the quotient rule. - Simplify the exponential terms in the numerator.

1. Apply the chain rule: \(h'(x)=\frac{e^x}{e^x+1}\). 2. Differentiate using the quotient rule: \(h''(x)=\frac{e^x(e^x+1)-e^x(e^x)}{(e^x+1)^2}\). 3. Simplify the numerator: \(h''(x)=\frac{e^x}{(e^x+1)^2}\).

\(h''(x)=\frac{e^x}{(e^x+1)^2}\)

- A degree-\(4\) polynomial becomes a constant after four derivatives. - What is the derivative of a constant? - Differentiate successively and track how the polynomial degree changes.

1. Differentiate twice: \(g'(x)=8x^3-6x\) and \(g''(x)=24x^2-6\). 2. Continue: \(g'''(x)=48x\) and \(g^{(4)}(x)=48\). 3. Differentiating the nonzero constant once more gives \(g^{(5)}(x)=0\). Therefore, the least such natural number is \(n=5\).

a) \(g''(x)=24x^2-6\) and \(g^{(4)}(x)=48\) b) \(n=5\)

- A graph's slope at a point is the value of its first derivative there. - The second derivative is the derivative of the first derivative. - For part c), work backward through two power-rule steps.

1. For part a), \(f'(x)=4ax^3\). The condition \(f'(1)=20\) gives \(4a=20\), so \(a=5\). 2. For part b), \(g'(x)=\frac{1}{2}x^2-10x\), so \(g''(x)=x-10\). 3. For part c), since \(\frac{d^2}{dx^2}(x^4)=12x^2\), one possible function is \(k(x)=x^4\). More generally, \(k(x)=x^4+Cx+D\).

a) \(a=5\) b) \(g''(x)=x-10\) c) One possible function is \(k(x)=x^4\).

- Each derivative decreases the exponent of a power function by \(1\). - Reverse two power-rule steps to recover the original exponent. - Match both the exponent and coefficient in \(an(n-1)x^{n-2}\). - Track the coefficient produced by differentiating twice.

1. For \(f(x)=ax^n\), the second derivative is \(f''(x)=an(n-1)x^{n-2}\). 2. In part a), \(n-2=1\), so \(n=3\). Then \(a(3)(2)=6\), giving \(a=1\). Thus \(f(x)=x^3\). 3. In part b), \(n-2=6\), so \(n=8\). Then \(a(8)(7)=56\), giving \(a=1\). Thus \(f(x)=x^8\). 4. In part c), \(n-2=k\), so \(n=k+2\). Matching coefficients gives \(a=1\). Thus \(f(x)=x^{k+2}\).

a) \(f(x)=x^3\) b) \(f(x)=x^8\) c) \(f(x)=x^{k+2}\)

- Differentiating twice decreases the exponent by \(2\). - Use \(f''(x)=an(n-1)x^{n-2}\) to find the coefficient \(a\). - Substitute the given input into the function you find.

1. Since the exponent in the second derivative is \(2\), \(n-2=2\), so \(n=4\). 2. For \(f(x)=ax^4\), \(f''(x)=12ax^2\). Matching coefficients gives \(12a=1.5\), so \(a=\frac{1}{8}\). 3. Therefore, \(f(x)=\frac{1}{8}x^4\). 4. Evaluate: \(f(2)=\frac{1}{8}(2)^4=2\).

1) \(f(x)=\frac{1}{8}x^4\) 2) \(f(2)=2\)

- Differentiate one step at a time and simplify the coefficients. - After how many derivatives does a degree-\(5\) power become constant? - Each derivative lowers the degree of a nonconstant polynomial by \(1\). - What are the derivatives of \(1\) and \(0\)?

1. Differentiate successively: \(g'(x)=\frac{1}{24}x^4\), \(g''(x)=\frac{1}{6}x^3\), \(g'''(x)=\frac{1}{2}x^2\), \(g^{(4)}(x)=x\), and \(g^{(5)}(x)=1\). 2. Since the derivative of \(1\) is \(0\), \(g^{(6)}(x)=0\). Every derivative of the zero function is also zero, so \(g^{(n)}(x)=0\) for all \(n\geq6\).

a) \(g'(x)=\frac{1}{24}x^4\), \(g''(x)=\frac{1}{6}x^3\), \(g'''(x)=\frac{1}{2}x^2\), \(g^{(4)}(x)=x\), and \(g^{(5)}(x)=1\) b) \(g^{(n)}(x)=0\) for every \(n\geq6\).

- Differentiate step by step and track how each coefficient is formed. - Leave the coefficient products unsimplified briefly to reveal a pattern. - What remains after differentiating \(x^n\) exactly \(n\) times?

1. Differentiate successively: \(f'(x)=5x^4\), \(f''(x)=20x^3\), \(f'''(x)=60x^2\), \(f^{(4)}(x)=120x\), and \(f^{(5)}(x)=120\). 2. The product \(5\cdot4\cdot3\cdot2\cdot1\) is \(120\), equal to the fifth derivative. 3. In general, the \(n\)th derivative of \(x^n\) is the constant \(n!=n(n-1)\cdots2\cdot1\).

1) \(f'(x)=5x^4\), \(f''(x)=20x^3\), \(f'''(x)=60x^2\), \(f^{(4)}(x)=120x\), and \(f^{(5)}(x)=120\) 2) Both values are \(120\). 3) \(g^{(n)}(x)=n!\)

- Simplify the original expression before differentiating. - Use the difference-of-squares identity. - Each derivative lowers the degree of a nonconstant polynomial by \(1\). - A polynomial becomes constant before its next derivative becomes zero.

1. Simplify: \(f(x)=(x^2-4)(x^2+4)=x^4-16\). 2. Differentiate successively: \(f'(x)=4x^3\), \(f''(x)=12x^2\), \(f'''(x)=24x\), \(f^{(4)}(x)=24\), and \(f^{(5)}(x)=0\). 3. Therefore, the least order is \(n=5\).

\(f'(x)=4x^3\) \(f''(x)=12x^2\) \(f'''(x)=24x\) \(f^{(4)}(x)=24\) \(f^{(5)}(x)=0\) The least order is \(n=5\).

- A term without \(x\) is constant. - The value \(\pi^2\) is a fixed number. - Test small-degree examples to identify the general pattern.

1. Since \(\pi^2\) is constant, the derivatives are \(g'(x)=\frac{1}{2}x^2-10x\), \(g''(x)=x-10\), \(g'''(x)=1\), and \(g^{(4)}(x)=0\). 2. Each derivative lowers a nonconstant polynomial's degree by \(1\). A degree-\(d\) polynomial has a nonzero constant \(d\)th derivative, so its \((d+1)\)th derivative is the first zero function.

a) \(g'(x)=\frac{1}{2}x^2-10x\), \(g''(x)=x-10\), \(g'''(x)=1\), and \(g^{(4)}(x)=0\) b) The \((d+1)\)th derivative is the first zero function.

- Use the product rule on the two factors. - Differentiate each factor before substituting into the rule. - Differentiate the simplified first derivative to find the second derivative. - Check signs and exponents while combining like terms.

1. Let \(u(x)=x^3-2x\) and \(v(x)=x^2+4\). Then \(u'(x)=3x^2-2\) and \(v'(x)=2x\). 2. Apply the product rule: \(f'(x)=(3x^2-2)(x^2+4)+(x^3-2x)(2x)\). 3. Simplify: \(f'(x)=5x^4+6x^2-8\). 4. Differentiate again: \(f''(x)=20x^3+12x\).

\(f'(x)=5x^4+6x^2-8\) \(f''(x)=20x^3+12x\)

- Apply the chain rule to the squared expression. - View the first derivative as a product when finding the second derivative. - Keep the factor from differentiating the outer square. - Expand only the expression needed to simplify the second derivative.

1. Apply the chain rule to the outer square and the inner function \(x^2-3x+5\): \(g'(x)=2(x^2-3x+5)(2x-3)\). 2. To find the second derivative, apply the product rule: \(g''(x)=2(2x-3)(2x-3)+2(x^2-3x+5)\cdot 2\). 3. Simplify: \(g''(x)=2(4x^2-12x+9)+4x^2-12x+20\) \(=12x^2-36x+38\).

\(g'(x)=2(x^2-3x+5)(2x-3)\) \(g''(x)=12x^2-36x+38\)

- Apply the quotient rule to find the first derivative. - Simplify the first derivative before differentiating again. - For the second derivative, write the denominator factor with a negative exponent and use the product and chain rules. - Combine the final terms over a common denominator.

1. Let \(u(x)=x^2-4\) and \(v(x)=x^2+4\). Then \(u'(x)=2x\) and \(v'(x)=2x\). 2. Apply the quotient rule: \(f'(x)=\frac{2x(x^2+4)-2x(x^2-4)}{(x^2+4)^2}=\frac{16x}{(x^2+4)^2}\). 3. Differentiate \(f'(x)=16x(x^2+4)^{-2}\) using the product and chain rules: \(f''(x)=16(x^2+4)^{-2}-64x^2(x^2+4)^{-3}\). 4. Write the terms over a common denominator and simplify: \(f''(x)=\frac{16(x^2+4)-64x^2}{(x^2+4)^3}=\frac{64-48x^2}{(x^2+4)^3}\).

\(f'(x)=\frac{16x}{(x^2+4)^2}\) \(f''(x)=\frac{64-48x^2}{(x^2+4)^3}\)

- Use the quotient rule for the first derivative. - Simplify before differentiating a second time. - When finding the second derivative, factor out a common \(x-1\) before expanding fully. - Keep track of the excluded value from the original denominator.

1. Apply the quotient rule to \(f\): \(f'(x)=\frac{3x^2(x-1)-x^3}{(x-1)^2}=\frac{2x^3-3x^2}{(x-1)^2}\). 2. Differentiate \(f'(x)=x^2(2x-3)(x-1)^{-2}\), or apply the quotient rule again. After combining terms, \(f''(x)=\frac{(6x^2-6x)(x-1)-2(2x^3-3x^2)}{(x-1)^3}\). 3. Simplify the numerator: \((6x^2-6x)(x-1)-2(2x^3-3x^2)=2x^3-6x^2+6x\). Thus, \(f''(x)=\frac{2x^3-6x^2+6x}{(x-1)^3}\). Both derivatives are defined for \(x\ne 1\).

\(f'(x)=\frac{2x^3-3x^2}{(x-1)^2}\) \(f''(x)=\frac{2x^3-6x^2+6x}{(x-1)^3}\) Both derivatives are defined for \(x\ne 1\).

- Rewrite the function using a negative exponent. - Use the chain rule for the first derivative. - Apply the product rule and chain rule when differentiating again.

1. Rewrite the function as \(f(x)=2(x^2+4)^{-1}\). By the chain rule, \(f'(x)=-2(x^2+4)^{-2}(2x)=-\frac{4x}{(x^2+4)^2}\). 2. Differentiate \(-4x(x^2+4)^{-2}\) using the product and chain rules: \(f''(x)=-4(x^2+4)^{-2}+(-4x)[-2(x^2+4)^{-3}(2x)]\) \(=-4(x^2+4)^{-2}+16x^2(x^2+4)^{-3}\). 3. Combine the terms: \(f''(x)=\frac{-4(x^2+4)+16x^2}{(x^2+4)^3}\) \(=\frac{12x^2-16}{(x^2+4)^3}\).

\(f'(x)=-\frac{4x}{(x^2+4)^2}\) \(f''(x)=\frac{12x^2-16}{(x^2+4)^3}\)

- Recall the derivative of cosine. - Multiply by the derivative of the inner function. - Use the product rule to find the second derivative. - Track the negative signs carefully.

1. By the chain rule, \(f'(x)=-\sin(3x^2+1)(6x)=-6x\sin(3x^2+1)\). 2. Differentiate the product using the product rule. The derivative of \(\sin(3x^2+1)\) is \(6x\cos(3x^2+1)\). Therefore, \(f''(x)=-6\sin(3x^2+1)-6x[6x\cos(3x^2+1)]\) \(=-6\sin(3x^2+1)-36x^2\cos(3x^2+1)\).

\(f'(x)=-6x\sin(3x^2+1)\) \(f''(x)=-6\sin(3x^2+1)-36x^2\cos(3x^2+1)\)

- Substitute the given point into the function. - Use the derivative to apply the slope condition. - Apply the chain rule each time you differentiate. - Compare the second derivative with the original function.

1. Since \(f(0)=a=5\), we have \(a=5\). 2. The first derivative is \(f'(x)=ake^{kx}\). The slope condition gives \(f'(0)=5k=-10\), so \(k=-2\). Thus \(f(x)=5e^{-2x}\). 3. Differentiate twice: \(f'(x)=-10e^{-2x}\) and \(f''(x)=20e^{-2x}\). 4. Since \(4f(x)=4(5e^{-2x})=20e^{-2x}\), it follows that \(f''(x)=4f(x)\).

1. \(a=5\), \(k=-2\) 2. \(f''(x)=20e^{-2x}\) 3. \(f''(x)=20e^{-2x}=4(5e^{-2x})=4f(x)\)

- Differentiate each exponential term separately. - Track the sign from differentiating \(e^{-3x}\). - Factor the second derivative to compare it with \(g\). - Use the effect of two successive derivatives to predict the fourth derivative.

1. Differentiate using the chain rule: \(g'(x)=3e^{3x}-3e^{-3x}\). Differentiating again gives \(g''(x)=9e^{3x}+9e^{-3x}\). 2. Factor out \(9\): \(g''(x)=9(e^{3x}+e^{-3x})=9g(x)\). 3. Applying two more derivatives introduces another factor of \(9\). Therefore, \(g^{(4)}(x)=9g''(x)=81g(x)\) \(=81e^{3x}+81e^{-3x}\).

1. \(g'(x)=3e^{3x}-3e^{-3x}\); \(g''(x)=9e^{3x}+9e^{-3x}\) 2. \(g''(x)=9g(x)\) 3. \(g^{(4)}(x)=81g(x)\)

- Differentiate each function twice. - Use the chain rule for every nontrivial exponent. - Track signs when the exponent contains \(-x\). - Constant terms disappear after differentiation.

1. For part a, \(f'(x)=2e^{4x-2}+3x^2\), so \(f''(x)=8e^{4x-2}+6x\). 2. For part b, \(g'(x)=e^{5-x}\), so \(g''(x)=-e^{5-x}\). 3. For part c, \(h'(x)=e^x-e^{-x}\), so \(h''(x)=e^x+e^{-x}\).

a) \(f''(x)=8e^{4x-2}+6x\) b) \(g''(x)=-e^{5-x}\) c) \(h''(x)=e^x+e^{-x}\)

- Find the second derivative before substituting the given x-value. - Apply the chain rule to the exponential function. - Set the resulting expression equal to the target value.

1. Differentiate twice: \(f'(x)=0.5ce^{0.5x}\) and \(f''(x)=0.25ce^{0.5x}\). 2. Substitute \(x=2\): \(f''(2)=0.25ce\). 3. Set \(0.25ce=e\). Since \(e\ne0\), \(0.25c=1\), so \(c=4\).

\(c=4\)

- Use the derivative rule for exponential functions with a constant base. - Differentiate each term separately. - Treat \(k\) as a constant. - Combine powers with the same base when useful.

1. Use \(\frac{d}{dx}a^x=(\ln a)a^x\). For part a, \(f'(x)=5(\ln 1.2)1.2^x\) and \(f''(x)=5(\ln 1.2)^2 1.2^x\). 2. For part b, \(g'(t)=0.5(\ln 4)4^t-2t\) and \(g''(t)=0.5(\ln 4)^2 4^t-2\). 3. Treat \(k\) as a positive constant. Since \(k^2k^x=k^{x+2}\), \(h'(x)=(\ln k)k^{x+2}\) and \(h''(x)=(\ln k)^2k^{x+2}\).

a) \(f'(x)=5(\ln 1.2)1.2^x\); \(f''(x)=5(\ln 1.2)^2 1.2^x\) b) \(g'(t)=0.5(\ln 4)4^t-2t\); \(g''(t)=0.5(\ln 4)^2 4^t-2\) c) \(h'(x)=(\ln k)k^{x+2}\); \(h''(x)=(\ln k)^2k^{x+2}\)

- Distinguish an exponential function from a power function. - Pull constant factors outside the derivative. - Use the chain rule for \(3^{-x}\). - The derivative of a constant is zero.

1. Differentiate the two terms in \(u\): \(u'(x)=\frac{(\ln 3)3^x-(\ln 3)3^{-x}}{\ln 3}=3^x-3^{-x}\). Therefore, \(u''(x)=(\ln 3)3^x+(\ln 3)3^{-x}\) \(=(\ln 3)(3^x+3^{-x})\). 2. In \(v\), \(\pi^x\) is exponential in \(x\), while \(x^\pi\) is a power of \(x\). Thus, \(v'(x)=(\ln\pi)\pi^x-\pi x^{\pi-1}\) and \(v''(x)=(\ln\pi)^2\pi^x-\pi(\pi-1)x^{\pi-2}\).

a) \(u'(x)=3^x-3^{-x}\); \(u''(x)=(\ln 3)(3^x+3^{-x})\) b) \(v'(x)=(\ln\pi)\pi^x-\pi x^{\pi-1}\); \(v''(x)=(\ln\pi)^2\pi^x-\pi(\pi-1)x^{\pi-2}\), for \(x>0\)

- Apply the chain rule to \(3^{2x}\). - Identify the repeated factor introduced by each derivative. - Write the nth derivative before solving the equation. - Cancel common nonzero factors and compare exponents.

1. By the chain rule, \(f'(x)=4(2\ln 3)3^{2x}=8(\ln 3)3^{2x}\). Differentiating again gives \(f''(x)=4(2\ln 3)^2 3^{2x}=16(\ln 3)^2 3^{2x}\). 2. Each differentiation introduces another factor of \(2\ln 3\). Therefore, \(f^{(n)}(x)=4(2\ln 3)^n3^{2x}\). 3. Substitute the formula and cancel the common nonzero factor \(4(2\ln 3)^n\): \(3^{2x}=81=3^4\). Hence, \(2x=4\), so \(x=2\).

a) \(f'(x)=8(\ln 3)3^{2x}\); \(f''(x)=16(\ln 3)^2 3^{2x}\) b) \(f^{(n)}(x)=4(2\ln 3)^n3^{2x}\) c) \(x=2\)

- Track the factor introduced by repeated differentiation of an exponential function. - Recall the nth derivative of \(x^n\). - Use the factorial definition to simplify coefficients. - Watch the alternating sign in repeated derivatives of \(e^{-x}\).

1. Each derivative of \(5^x\) introduces a factor of \(\ln 5\). After \(n\) derivatives, \(g^{(n)}(x)=\frac{(\ln 5)^n5^x}{(\ln 5)^n}=5^x\). 2. The nth derivative of \(e^{4x}\) is \(4^ne^{4x}\), and the nth derivative of \(x^n\) is \(n!\). Thus, \(h^{(n)}(x)=4^ne^{4x}+n!\). 3. The nth derivative of \(\frac{x^n}{n!}\) is \(1\). Each derivative of \(e^{-x}\) changes the sign, so \(k^{(n)}(x)=1+(-1)^ne^{-x}\).

a) \(g^{(n)}(x)=5^x\) b) \(h^{(n)}(x)=4^ne^{4x}+n!\) c) \(k^{(n)}(x)=1+(-1)^ne^{-x}\)

- Use the derivative of the natural logarithm. - Include the inner derivative for a linear logarithm input. - Rewrite a logarithm with base \(3\) using natural logarithms. - Treat \(a\) as a fixed constant.

1. For part a, apply the chain rule twice: \(f'(x)=\frac{5}{5x-3}\) and \(f''(x)=-\frac{25}{(5x-3)^2}\). 2. Rewrite \(\log_3x=\frac{\ln x}{\ln3}\). Then \(g'(x)=\frac{1}{x\ln3}-4\) and \(g''(x)=-\frac{1}{x^2\ln3}\). 3. Treat \(a\) as a constant: \(h'(t)=\frac{a}{at+1}\) and \(h''(t)=-\frac{a^2}{(at+1)^2}\).

a) \(f'(x)=\frac{5}{5x-3}\), \(f''(x)=-\frac{25}{(5x-3)^2}\) b) \(g'(x)=\frac{1}{x\ln3}-4\), \(g''(x)=-\frac{1}{x^2\ln3}\) c) \(h'(t)=\frac{a}{at+1}\), \(h''(t)=-\frac{a^2}{(at+1)^2}\)

- Apply the product rule in part b. - The first derivatives in parts a and c are quotients. - Include inner derivatives for logarithmic composites. - Simplify before differentiating a second time.

1. For part a, \(f'(x)=\frac{2x}{x^2+1}\). Using the quotient rule, \(f''(x)=\frac{2-2x^2}{(x^2+1)^2}\). 2. For part b, apply the product rule: \(g'(x)=2x\ln x+x\). Differentiating again gives \(g''(x)=2\ln x+3\). 3. For part c, apply the chain rule: \(h'(x)=-\frac{\sin(\ln x)}{x}\). Differentiating again gives \(h''(x)=\frac{\sin(\ln x)-\cos(\ln x)}{x^2}\).

a) \(f'(x)=\frac{2x}{x^2+1}\), \(f''(x)=\frac{2-2x^2}{(x^2+1)^2}\) b) \(g'(x)=2x\ln x+x\), \(g''(x)=2\ln x+3\) c) \(h'(x)=-\frac{\sin(\ln x)}{x}\), \(h''(x)=\frac{\sin(\ln x)-\cos(\ln x)}{x^2}\)

- Split the fraction into separate terms before differentiating. - Rewrite reciprocal exponentials with negative exponents. - Differentiate each term separately. - Apply the chain rule to \(e^{-x}\).

1. Simplify the original function: \(f(x)=e^x-4e^{-x}+x^2\). 2. Differentiate term by term: \(f'(x)=e^x+4e^{-x}+2x\). 3. Differentiate again: \(f''(x)=e^x-4e^{-x}+2\).

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

- Use the chain rule for the first derivative. - The first derivative is a product, so use the product rule next. - Apply the chain rule again to the exponential factor. - Factor out the common exponential when simplifying.

1. The derivative of the exponent \(5-2x^2\) is \(-4x\). By the chain rule, \(g'(x)=-4xe^{5-2x^2}\). 2. Differentiate the product using the product and chain rules: \(g''(x)=-4e^{5-2x^2}+(-4x)(-4x)e^{5-2x^2}\). 3. Factor the common exponential: \(g''(x)=(16x^2-4)e^{5-2x^2}\).

\(g'(x)=-4xe^{5-2x^2}\) \(g''(x)=(16x^2-4)e^{5-2x^2}\)

- Use the product rule for each derivative. - Apply the chain rule to the exponential factor. - Factor out the exponential term after each differentiation. - Watch the signs when differentiating \(e^{-x}\).

1. For a), apply the product rule: \(f'(x)=-2e^{2x}+(5-2x)(2e^{2x})\) \(=(8-4x)e^{2x}\). Differentiate again: \(f''(x)=-4e^{2x}+(8-4x)(2e^{2x})\) \(=(12-8x)e^{2x}\). 2. For b), \(f'(x)=(2x+4)e^{-x}-(x^2+4x-1)e^{-x}\) \(=(-x^2-2x+5)e^{-x}\). Differentiate again: \(f''(x)=(-2x-2)e^{-x}-(-x^2-2x+5)e^{-x}\) \(=(x^2-7)e^{-x}\).

a) \(f'(x)=(8-4x)e^{2x}\) and \(f''(x)=(12-8x)e^{2x}\) b) \(f'(x)=(-x^2-2x+5)e^{-x}\) and \(f''(x)=(x^2-7)e^{-x}\)

- Use the derivative rule for \(a^x\). - Apply the chain rule to the exponent \(2x\). - Use \(\ln(0.5)=-\ln 2\). - Differentiate the first derivative term by term.

1. Differentiate the first term using the exponential rule and the chain rule: \(\frac{d}{dx}[4\cdot 3^{2x}]=8(\ln 3)3^{2x}\). For the second term, \(\frac{d}{dx}[-0.5^x]=-(\ln 0.5)0.5^x=(\ln 2)0.5^x\). Therefore, \(f'(x)=8(\ln 3)3^{2x}+(\ln 2)0.5^x\). 2. Differentiate again: \(f''(x)=16(\ln 3)^2 3^{2x}-(\ln 2)^2 0.5^x\).

\(f'(x)=8(\ln 3)3^{2x}+(\ln 2)0.5^x\) \(f''(x)=16(\ln 3)^2 3^{2x}-(\ln 2)^2 0.5^x\)

- Differentiate each term separately. - Use the chain rule for exponents that depend on \(x\). - The constant term disappears. - Track every factor introduced by the exponent \(3x\).

1. Differentiate each term: \(f'(x)=(\ln 5)5^{x-1}+6(\ln 0.1)0.1^{3x}-2x^2\). 2. Differentiate again. The chain rule introduces another factor of \(3\) in the second exponential term: \(f''(x)=(\ln 5)^2 5^{x-1}+18(\ln 0.1)^2 0.1^{3x}-4x\).

\(f'(x)=(\ln 5)5^{x-1}+6(\ln 0.1)0.1^{3x}-2x^2\) \(f''(x)=(\ln 5)^2 5^{x-1}+18(\ln 0.1)^2 0.1^{3x}-4x\)

- Identify the product and composite functions before differentiating. - Use the double-angle identity \(2\sin x\cos x=\sin(2x)\). - Include the inner derivative when differentiating \(\cos(2x)\) or \(\cos^2x\).

1. The linear and trigonometric expressions are defined for every real input, so \(D_f=\mathbb{R}\). 2. Differentiate using the product rule and chain rule: \(f'(x)=\sin(2x)+2x\cos(2x)-2\sin x\cos x\). Since \(2\sin x\cos x=\sin(2x)\), this simplifies to \(f'(x)=2x\cos(2x)\). 3. Differentiate again using the product rule: \(f''(x)=2\cos(2x)-4x\sin(2x)\).

\(D_f=\mathbb{R}\) \(f'(x)=2x\cos(2x)\) \(f''(x)=2\cos(2x)-4x\sin(2x)\)

- Recall that sine is odd and cosine is even. - Apply the product rule to find the first derivative. - Use the chain rule when differentiating a function evaluated at \(-x\). - Track the parity after each differentiation.

1. Evaluate \(f(-x)\): \(f(-x)=(( -x)^2-4)\sin(-x)=-(x^2-4)\sin(x)=-f(x)\). Therefore, \(f\) is odd and its graph has origin symmetry. 2. By the product rule, \(f'(x)=2x\sin(x)+(x^2-4)\cos(x)\). Then \(f'(-x)=2(-x)\sin(-x)+(( -x)^2-4)\cos(-x)=2x\sin(x)+(x^2-4)\cos(x)=f'(x)\). Thus \(f'\) is even and its graph has y-axis symmetry. 3. Suppose \(f(-x)=-f(x)\). Differentiate both sides with respect to \(x\): \(-f'(-x)=-f'(x)\), so \(f'(-x)=f'(x)\). Differentiate again: \(-f''(-x)=f''(x)\), so \(f''(-x)=-f''(x)\). Therefore, the second derivative is odd and has origin symmetry.

1) \(f(-x)=-f(x)\), so the graph has origin symmetry. 2) \(f'(x)=2x\sin(x)+(x^2-4)\cos(x)\), and \(f'(-x)=f'(x)\), so its graph has y-axis symmetry. 3) Differentiating \(f(-x)=-f(x)\) twice gives \(f''(-x)=-f''(x)\), so the second derivative has origin symmetry.

- Apply the chain rule at each differentiation step. - Compare the second and fourth derivatives with the original function. - Substitute the derivatives into the given equation and factor out \(\cos(kx)\). - Use a substitution for the equation containing only even powers of \(k\).

1. Differentiate repeatedly using the chain rule: \(f_k'(x)=-k\sin(kx)\), \(f_k^{(2)}(x)=-k^2\cos(kx)\), \(f_k^{(3)}(x)=k^3\sin(kx)\), and \(f_k^{(4)}(x)=k^4\cos(kx)\). 2. Substitute into the differential equation: \(k^4\cos(kx)-20k^2\cos(kx)=-64\cos(kx)\). Therefore, \((k^4-20k^2+64)\cos(kx)=0\) for every real \(x\). Since \(\cos(kx)\) is not identically zero, \(k^4-20k^2+64=0\). Let \(u=k^2\). Then \(u^2-20u+64=(u-16)(u-4)=0\). Thus, \(k^2=16\) or \(k^2=4\), giving \(k\in\{-4, -2, 2, 4\}\).

1. \(f_k^{(4)}(x)=k^4\cos(kx)\) 2. \(k\in\{-4, -2, 2, 4\}\)

- The tangent line requires the function value and derivative value at the specified input. - Recall the derivative cycle for sine and cosine. - Track the signs during repeated differentiation. - Compare the fourth derivative with the original function.

1. The first derivative is \(f'(x)=\cos x\). At \(x_0=\frac{\pi}{3}\), \(f\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}\) and \(f'\left(\frac{\pi}{3}\right)=\frac{1}{2}\). Therefore, the tangent line is \(t(x)=\frac{1}{2}\left(x-\frac{\pi}{3}\right)+\frac{\sqrt{3}}{2}\), or equivalently, \(t(x)=\frac{1}{2}x-\frac{\pi}{6}+\frac{\sqrt{3}}{2}\). 2. Repeated differentiation gives \(f'(x)=\cos x\), \(f^{(2)}(x)=-\sin x\), \(f^{(3)}(x)=-\cos x\), and \(f^{(4)}(x)=\sin x\). Thus, \(f^{(4)}(x)=f(x)\).

a) \(t(x)=\frac{1}{2}x-\frac{\pi}{6}+\frac{\sqrt{3}}{2}\) b) \(f'(x)=\cos x\), \(f^{(2)}(x)=-\sin x\), \(f^{(3)}(x)=-\cos x\), and \(f^{(4)}(x)=\sin x\). Therefore, \(f^{(4)}=f\).

- List the derivative cycle of cosine. - Use the remainder when \(15\) is divided by \(4\). - Solve the resulting trigonometric equation using the unit circle.

1. The derivatives of cosine repeat in a cycle of length \(4\): \(\cos x,-\sin x,-\cos x,\sin x,\cos x\). 2. Since \(15=3\cdot4+3\), the 15th derivative matches the third derivative in the cycle: \(g^{(15)}(x)=\sin x\). 3. Solve \(\sin x=\frac{1}{2}\) on \([0, 2\pi]\). The solutions are \(x=\frac{\pi}{6}\) and \(x=\frac{5\pi}{6}\).

\(x=\frac{\pi}{6}\) and \(x=\frac{5\pi}{6}\)

- Use the definition of an even function. - Apply the chain rule to \(e^{-x}\). - Set the first derivative equal to zero. - Use the sign of the second derivative to classify the critical point.

1. Evaluate \(f(-x)\): \(f(-x)=\frac{1}{2}(e^{-x}+e^x)=f(x)\). Therefore, \(f\) is even and its graph has y-axis symmetry. 2. Differentiate: \(f'(x)=\frac{1}{2}(e^x-e^{-x})\) and \(f''(x)=\frac{1}{2}(e^x+e^{-x})\). Thus \(f''(x)=f(x)\). 3. Solve \(f'(x)=0\): \(e^x=e^{-x}\), so \(e^{2x}=1\) and \(x=0\). Since \(f''(0)=1>0\), the point is a local minimum. Its coordinates are \((0,1)\).

1) \(f(-x)=f(x)\), so the graph has y-axis symmetry. 2) \(f'(x)=\frac{1}{2}(e^x-e^{-x})\), \(f''(x)=\frac{1}{2}(e^x+e^{-x})\), and \(f''=f\). 3) Local minimum at \((0,1)\).

- Zeros of the first derivative occur where the graph has horizontal tangent lines. - The sign of the first derivative indicates whether the function is increasing or decreasing. - Zeros of the second derivative can occur where the graph changes concavity. - Recall how the derivative of an even function behaves.

1. Statement (1) is true. The graph of \(f\) has horizontal tangent lines at its local maxima \(x=-2\) and \(x=2\) and at its local minimum \(x=0\). Thus, \(f^{\prime}\) has exactly three zeros. 2. Statement (2) is false. The function is decreasing for \(x>2\), so \(f^{\prime}(x)<0\) there. 3. Statement (3) is true. The graph changes concavity twice, near \(x=-1.15\) and \(x=1.15\), so \(f^{\prime\prime}\) has exactly two zeros on the interval. 4. Statement (4) is false. The graph of \(f\) is symmetric about the y-axis, so \(f\) is even and \(f^{\prime}\) is odd. Therefore, the graph of \(f^{\prime}\) is symmetric about the origin. 5. Statement (5) is true. The graph of \(f\) is increasing at \(x=1\), so \(f^{\prime}(1)>0\).

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

- Match local extrema of one graph with zeros of another. - Compare increasing and decreasing behavior with the sign of the derivative. - After matching one derivative pair, check whether the remaining graph is the derivative of the middle graph.

1. Graph \(a\) has a local maximum at \(x = -2\) and a local minimum at \(x = 2\). Graph \(b\) is zero at those same values, is negative between them, and is positive outside them. Therefore, \(b\) is the derivative of \(a\). 2. Graph \(b\) has a local minimum at \(x = 0\). Graph \(c\) is zero at \(x = 0\), is negative for \(x < 0\), and is positive for \(x > 0\). Therefore, \(c\) is the derivative of \(b\). 3. Thus, \(a\) is the original function, \(b\) is its first derivative, and \(c\) is its second derivative.

Graph \(a\) is the original function, graph \(b\) is the first derivative, and graph \(c\) is the second derivative.

- Compare where the blue graph increases or decreases with the sign of the green graph. - The value of the derivative gives the slope of the original function. - The second derivative is the slope of the first-derivative graph.

1. Graph p increases for \(x<3\) and decreases for \(x>3\). Graph q is positive for \(x<3\), zero at \(x=3\), and negative for \(x>3\). Therefore, graph p is \(r\), and graph q is \(s=r^{\prime}\). 2. The slope of \(r\) at \(x=1\) is \(r^{\prime}(1)=s(1)=2\). 3. The second derivative \(r^{\prime\prime}(1)\) is the slope of the line \(s\). Using \((0,3)\) and \((3,0)\), the slope is \(\frac{0-3}{3-0}=-1\).

a) Graph p is \(r\), and graph q is \(s=r^{\prime}\). b) \(r^{\prime}(1)=2\) c) \(r^{\prime\prime}(1)=-1\)

- Use the sign change of \(f^{\prime}\) at \(x=0\). - A function decreases where its derivative is negative. - The second derivative is the slope of the first-derivative graph.

1. At \(x=0\), \(f^{\prime}\) changes from positive to negative, so \(f\) changes from increasing to decreasing and has a local maximum. Statement a) is true. 2. For \(0<x<2\), \(f^{\prime}(x)<0\), so \(f\) is strictly decreasing on \([0,2]\). Statement b) is true. 3. The value \(f^{\prime\prime}(1)\) is the slope of the graph of \(f^{\prime}\) at \(x=1\). The derivative graph is decreasing there, so \(f^{\prime\prime}(1)<0\). Statement c) is true.

a) True b) True c) True

- Count all horizontal tangent lines. - Count all changes in concavity.

1. The graph has one local maximum at \(x=0\). The tangent line is horizontal there, so \(h^{\prime}\) has exactly \(1\) zero. 2. The graph changes concavity once on each side of the maximum. Therefore, it has two inflection points and \(h^{\prime\prime}\) has exactly \(2\) zeros.

\(h^{\prime}\) has \(1\) zero, and \(h^{\prime\prime}\) has \(2\) zeros.

- Count points with horizontal tangent lines. - Check for a concavity change within each continuous branch. - Do not count a vertical asymptote as an inflection point.

1. The graph has one local maximum on the left branch and one local minimum on the right branch. Therefore, \(p^{\prime}\) has \(2\) zeros. 2. Neither branch changes concavity. The function is not defined at the vertical asymptote, so no change across it is an inflection point. Therefore, \(p^{\prime\prime}\) has \(0\) zeros.

\(p^{\prime}\) has \(2\) zeros, and \(p^{\prime\prime}\) has \(0\) zeros.

- Look for horizontal tangent lines. - Identify any point where a continuous branch changes concavity. - Analyze the separate branches around the vertical asymptotes.

1. The function is strictly decreasing on every interval of its domain and has no horizontal tangent lines. Therefore, \(r^{\prime}\) has \(0\) zeros. 2. On the middle branch, the graph changes concavity at the origin. Thus, \((0,0)\) is an inflection point and \(r^{\prime\prime}\) has one zero. The outer branches have no additional concavity changes.

\(r^{\prime}\) has \(0\) zeros, and \(r^{\prime\prime}\) has \(1\) zero.

- Differentiation produces two terms with the same exponential factor. - Compare the linear factor before and after differentiating.

1. Base case: For \(n=0\), \(f^{(0)}(x)=f(x)=(x+0)e^x\). 2. Inductive hypothesis: Assume \(f^{(k)}(x)=(x+k)e^x\) for an arbitrary integer \(k\ge0\). 3. Using the product rule, \(f^{(k+1)}(x)=e^x+(x+k)e^x=(x+k+1)e^x\). 4. This is the formula for \(k+1\), so it holds for every integer \(n\ge0\).

For every integer \(n\ge0\), \(f^{(n)}(x)=(x+n)e^x\).

- Use the product rule for the product of a trigonometric function and \(g(x)\). - Recall the derivatives of sine and cosine. - When a factor is already a derivative, such as \(g'(x)\), differentiating raises its order by one. - For the second derivative in b), use the chain rule on \((g'(x))^2\).

1. For a), apply the product rule: \(f'(x) = \cos(x)g(x) + \sin(x)g'(x)\). 2. Differentiate both terms again: \(f''(x) = -\sin(x)g(x) + \cos(x)g'(x) + \cos(x)g'(x) + \sin(x)g''(x)\) \(= -\sin(x)g(x) + 2\cos(x)g'(x) + \sin(x)g''(x)\). 3. For b), apply the product rule: \(f'(x) = (g'(x))^2 + g(x)g''(x)\). 4. Differentiate again. By the chain rule, \(\frac{d}{dx}[(g'(x))^2] = 2g'(x)g''(x)\), and by the product rule, \(\frac{d}{dx}[g(x)g''(x)] = g'(x)g''(x) + g(x)g'''(x)\). Therefore, \(f''(x) = 3g'(x)g''(x) + g(x)g'''(x)\).

a) \(f'(x) = \cos(x)g(x) + \sin(x)g'(x)\) and \(f''(x) = -\sin(x)g(x) + 2\cos(x)g'(x) + \sin(x)g''(x)\) b) \(f'(x) = (g'(x))^2 + g(x)g''(x)\) and \(f''(x) = 3g'(x)g''(x) + g(x)g'''(x)\)

- For three factors, differentiate one factor at a time while leaving the other two unchanged. - Differentiate each term of the first derivative separately to obtain the second derivative. - In part b), factor out the common term \(e^x\) before combining like trigonometric terms.

1. First differentiate the product of three functions: \(f'(x) = u'vw + uv'w + uvw'\). 2. Differentiate each term: \((u'vw)' = u''vw + u'v'w + u'vw'\), \((uv'w)' = u'v'w + uv''w + uv'w'\), \((uvw')' = u'vw' + uv'w' + uvw''\). 3. Combine like terms: \(f'' = u''vw + uv''w + uvw'' + 2(u'v'w + u'vw' + uv'w')\). 4. For \(u=x^2\), \(v=e^x\), and \(w=\sin(x)\), \(u'=2x\), \(u''=2\), \(v'=v''=e^x\), \(w'=\cos(x)\), and \(w''=-\sin(x)\). 5. Substitute into the formula: \(f''(x) = 2e^x\sin(x) + x^2e^x\sin(x) - x^2e^x\sin(x)\) \(+ 2[2xe^x\sin(x) + 2xe^x\cos(x) + x^2e^x\cos(x)]\). 6. Factor and combine terms: \(f''(x) = e^x[(2+4x)\sin(x) + (4x+2x^2)\cos(x)]\).

a) \(f'' = u''vw + uv''w + uvw'' + 2(u'v'w + u'vw' + uv'w')\) b) \(f''(x) = e^x[(2+4x)\sin(x) + (4x+2x^2)\cos(x)]\)

- Apply the chain rule and include the derivative of \(1-kx\). - Track how the exponent changes after each derivative. - What happens when a degree-12 polynomial is differentiated 12 times? - Count how many factors of \(-k\) appear.

1. Apply the chain rule: \(g_k'(x)=12(1-kx)^{11}(-k)=-12k(1-kx)^{11}\). 2. Differentiate again: \(g_k''(x)=-12k\cdot 11(1-kx)^{10}(-k)\) \(=132k^2(1-kx)^{10}\). 3. Each derivative lowers the exponent by \(1\), multiplies by the current exponent, and contributes another factor of \(-k\). Therefore, \(g_k^{(12)}(x)=12!(-k)^{12}\). Since \(12\) is even and \(12!=479{,}001{,}600\), \(g_k^{(12)}(x)=479{,}001{,}600k^{12}\).

\(g_k''(x)=132k^2(1-kx)^{10}\) \(g_k^{(12)}(x)=479{,}001{,}600k^{12}\)

- Use the product rule for the first few derivatives. - Compare the linear factor in each result. - Identify what stays fixed and what changes with the derivative order. - Write a formula in terms of \(n\) before substituting \(100\).

1. Apply the product rule repeatedly: \(g'(x)=2e^x+(2x-1)e^x=(2x+1)e^x\), \(g''(x)=2e^x+(2x+1)e^x=(2x+3)e^x\), \(g'''(x)=2e^x+(2x+3)e^x=(2x+5)e^x\). 2. Each derivative increases the constant in the linear factor by \(2\). The pattern is \(g^{(n)}(x)=(2x+2n-1)e^x\). 3. Substitute \(n=100\): \(g^{(100)}(x)=(2x+200-1)e^x=(2x+199)e^x\).

\(g^{(100)}(x)=(2x+199)e^x\)

- Use the quotient rule or negative-exponent form to differentiate. - Clear the denominators after canceling the nonzero parameter. - Use a substitution for the equation containing only even powers.

1. Differentiating twice gives \(g_a'(x)=\frac{-2ax}{(x^2+1)^2}\) and \(g_a''(x)=\frac{a(6x^2-2)}{(x^2+1)^3}\). 2. At an intersection, \(\frac{a}{x^2+1}=\frac{a(6x^2-2)}{(x^2+1)^3}\). 3. Since \(a\ne0\) and \(x^2+1>0\), simplify to \((x^2+1)^2=6x^2-2\), or \(x^4-4x^2+3=0\). 4. Let \(u=x^2\). Then \(u^2-4u+3=(u-1)(u-3)=0\), so \(x^2=1\) or \(x^2=3\). 5. Therefore, \(x=\pm1\) or \(x=\pm\sqrt{3}\), independent of \(a\).

\(x\in\{-\sqrt{3}, -1, 1, \sqrt{3}\}\)

- Rewrite the rational term using a negative exponent. - Analyze signs separately on the two sides of the excluded value. - Track how the exponent changes with each derivative. - A vertical asymptote occurs when function magnitude grows without bound near a point.

1. The first derivative is \(g'(x)=\frac{6}{(x-4)^3}\). It is negative for \(x<4\) and positive for \(x>4\). Thus \(g\) is strictly decreasing on \((-\infty,4)\) and strictly increasing on \((4,\infty)\). 2. The second derivative is \(g''(x)=\frac{-18}{(x-4)^4}<0\) throughout the domain. Therefore, the graph is concave down on \((-\infty,4)\) and \((4,\infty)\). 3. Repeated differentiation gives \(g^{(n)}(x)=a_n(x-4)^{-(n+2)}=\frac{a_n}{(x-4)^{n+2}}\), where \(a_n\neq0\). As \(x\to4\), the denominator approaches \(0\) while the numerator remains nonzero, so \(|g^{(n)}(x)|\to\infty\). Therefore, \(x=4\) is a vertical asymptote of every positive-order derivative.

1. Strictly decreasing on \((-\infty,4)\); strictly increasing on \((4,\infty)\). 2. Concave down on \((-\infty,4)\) and \((4,\infty)\). 3. For every positive integer \(n\), \(g^{(n)}(x)=\frac{a_n}{(x-4)^{n+2}\!}\) with \(a_n\neq0\), so \(x=4\) is a vertical asymptote.

- Relate a sign change across a pole to whether its pole order is odd or even. - Think about what differentiating a negative power does to its exponent. - A single counterexample is enough to disprove an “always” statement.

1. Near a pole of order \(k\), the dominant factor of \(f\) is \(C(x-x_0)^{-k}\). Differentiating this factor gives \(-kC(x-x_0)^{-(k+1)}\), so the derivative has pole order \(k+1\). 2. Sarah’s statement is true. If \(f\) changes sign across the pole, then \(k\) is odd. Therefore, \(k+1\) is even, so the dominant factor of \(f^{\prime}\) has the same sign on both sides of \(x_0\). Thus, \(f^{\prime}\) does not change sign across the asymptote sufficiently close to \(x_0\). 3. Leo’s statement is false. For example, \(f(x)=\frac{1}{x}\) changes sign across \(x=0\), while \(f^{\prime}(x)=-\frac{1}{x^2}\) is negative on both sides of \(0\).

Sarah’s statement is true for reduced rational functions: an odd-order pole of \(f\) becomes an even-order pole of \(f^{\prime}\). Leo’s statement is false. A counterexample is \(f(x)=\frac{1}{x}\), for which \(f\) changes sign across \(0\) but \(f^{\prime}(x)=-\frac{1}{x^2}\) does not.

- Differentiate in stages and combine terms over a common denominator. - Squaring can remove the power \(3/2\), but verify all resulting positive solutions. - Use the function value and derivative value to write the tangent equation.

1. Differentiate: \(g_a'(x)=\frac{2x}{\sqrt{2x^2+a}}\). 2. Differentiate again and simplify: \(g_a''(x)=\frac{2a}{(2x^2+a)^{3/2}}\). 3. Set \(\frac{2a}{(a+2)^{3/2}}=\frac{1}{2}\). Squaring the positive quantities gives \(16a^2=(a+2)^3\). 4. Rearranging and factoring gives \((a-2)(a^2-8a-4)=0\). The positive solutions are \(a=2\) and \(a=4+2\sqrt{5}\). 5. For \(a=1\), \(g_1(2)=3\) and \(g_1'(2)=\frac{4}{3}\). 6. The tangent line is \(y-3=\frac{4}{3}(x-2)\), or \(y=\frac{4}{3}x+\frac{1}{3}\).

a) \(g_a''(x)=\frac{2a}{(2x^2+a)^{3/2}}\) b) \(a=2\) or \(a=4+2\sqrt{5}\) c) \(y=\frac{4}{3}x+\frac{1}{3}\)

- Find where the original denominator equals zero. - After applying the quotient rule, use \(\sin^2x+\cos^2x=1\). - Rewrite \((\sin x+\cos x)^2\) using a double-angle identity. - Differentiate the compact first-derivative form using the chain rule.

1. The denominator is zero when \(\sin x+\cos x=0\). On \([0, 2\pi]\), this occurs at \(x=\frac{3\pi}{4}\) and \(x=\frac{7\pi}{4}\). Therefore, \(D_f=[0, 2\pi]\setminus\left\{\frac{3\pi}{4},\frac{7\pi}{4}\right\}\). 2. Apply the quotient rule: \(f'(x)=\frac{(\cos x+\sin x)(\sin x+\cos x)-(\sin x-\cos x)(\cos x-\sin x)}{(\sin x+\cos x)^2}\). The numerator simplifies to \((\sin x+\cos x)^2+(\sin x-\cos x)^2=2\). Also, \((\sin x+\cos x)^2=1+\sin(2x)\). Hence, \(f'(x)=\frac{2}{1+\sin(2x)}\). 3. Differentiate using the chain rule: \(f''(x)=-2(1+\sin(2x))^{-2}\cdot 2\cos(2x)\) \(=\frac{-4\cos(2x)}{(1+\sin(2x))^2}\).

1. \(D_f=[0, 2\pi]\setminus\left\{\frac{3\pi}{4},\frac{7\pi}{4}\right\}\) 2. \(f'(x)=\frac{2}{1+\sin(2x)}\) \(f''(x)=\frac{-4\cos(2x)}{(1+\sin(2x))^2}\)

- Differentiate once and solve the critical-point equation. - Eliminate \(k\) from the coordinates of the extremum. - Compute the first few derivatives and look for a pattern. - Use mathematical induction to justify the nth-derivative formula.

1. Differentiate: \(g_k'(x)=(kx+k+1)e^x\). The critical number satisfies \(kx+k+1=0\), so \(x=-1-\frac1k\). At this x-coordinate, \(y=-ke^{-1-1/k}\). Also, \(g_k''\left(-1-\frac1k\right)=ke^{-1-1/k}\). Therefore, the point is a local minimum when \(k>0\) and a local maximum when \(k<0\). 2. From \(x=-1-\frac1k\), solve for the parameter: \(k=-\frac1{x+1}\). Since the extremum value can be written as \(y=-ke^x\), substitution gives \(y=\frac{e^x}{x+1}\), with \(x\ne-1\). 3. For every integer \(n\ge0\), \(g_k^{(n)}(x)=(kx+nk+1)e^x\). The formula is true for \(n=0\). If it is true for some \(n\), then the product rule gives \(g_k^{(n+1)}(x)=\left(k+kx+nk+1\right)e^x=(kx+(n+1)k+1)e^x\). Therefore, the formula follows by induction.

a) \(E_k=\left(-1-\frac1k, -ke^{-1-1/k}\right)\). It is a local minimum for \(k>0\) and a local maximum for \(k<0\). b) \(y=\frac{e^x}{x+1}\), with \(x\ne-1\) c) \(g_k^{(n)}(x)=(kx+nk+1)e^x\), for integers \(n\ge0\).

- Differentiate using the product and chain rules, or rewrite the function with a negative exponent. - Substitute \(z=x^2\) after setting the function equal to its second derivative. - Relate positive solutions in \(z\) to pairs of real solutions in \(x\). - Use the discriminant and the signs of the roots of the quadratic in \(z\).

1. Differentiate twice: \(g_a'(x)=-2ax(x^2+a)^{-2}\), so \(g_a''(x)=-2a(x^2+a)^{-2}+8ax^2(x^2+a)^{-3}\) \(=\frac{6ax^2-2a^2}{(x^2+a)^3}\). 2. Because \(a>0\), the denominators are positive. Setting \(g_a(x)=g_a''(x)\) and simplifying gives \((x^2+a)^2=6x^2-2a\). Let \(z=x^2\), where \(z\ge0\). Then \(z^2+(2a-6)z+a^2+2a=0\). Its discriminant is \(\Delta=(2a-6)^2-4(a^2+2a)=36-32a\). If \(0<a<\frac98\), the quadratic has two distinct roots. Their product is \(a^2+2a>0\), and their sum is \(6-2a>0\), so both roots are positive. This gives four real values of \(x\). If \(a=\frac98\), the quadratic has the repeated positive root \(z=3-a=\frac{15}{8}\), which gives exactly two real solutions, \(x=\pm\sqrt{\frac{15}{8}}\). If \(a>\frac98\), there are no real roots in \(z\). Therefore, the required value is \(a=\frac98\). 3. The inequality is equivalent to \(q_a(z)=z^2+(2a-6)z+a^2+2a\ge0\) for every \(z\ge0\). If \(a\ge\frac98\), then \(\Delta\le0\), and the upward-opening quadratic is nonnegative for all real \(z\). If \(0<a<\frac98\), it has two positive roots and is negative between them. Therefore, \(a\ge\frac98\).

1) \(g_a''(x)=\frac{6ax^2-2a^2}{(x^2+a)^3}\) 2) \(a=\frac98\) 3) \(a\ge\frac98\)

- Zeros of the first derivative occur where the graph has horizontal tangent lines. - Zeros of the second derivative can occur at inflection points. - A change across a discontinuity is not an inflection point. - Count each qualifying point within the function domain.

1. For \(f\), zeros of \(f^{\prime}\) occur at horizontal tangent lines. The graph has one local minimum at \(x=-1\) and one local maximum at \(x=1\), so \(f^{\prime}\) has \(2\) zeros. Zeros of \(f^{\prime\prime}\) correspond to inflection points. The graph changes concavity three times: once on the left side, once at the origin, and once on the right side. Therefore, \(f^{\prime\prime}\) has \(3\) zeros. 2. For \(g\), the graph has one local minimum at \(x=0\), so \(g^{\prime}\) has \(1\) zero. The graph does not change concavity within any interval of its domain. Changes across the vertical asymptotes \(x=-1\) and \(x=1\) are not inflection points because the function is not defined there. Therefore, \(g^{\prime\prime}\) has \(0\) zeros.

a) \(f^{\prime}\) has \(2\) zeros, and \(f^{\prime\prime}\) has \(3\) zeros. b) \(g^{\prime}\) has \(1\) zero, and \(g^{\prime\prime}\) has \(0\) zeros.

- Zeros of the first derivative occur at horizontal tangent lines. - Zeros of the second derivative can occur where the graph changes concavity. - Do not count a change across a vertical asymptote as an inflection point. - Track each branch from left to right.

1. In panel a), the graph of \(f\) has one local minimum at \(x=0\), so \(f^{\prime}\) has \(1\) zero. Within each continuous interval of its domain, the graph does not change concavity. The changes across the vertical asymptotes are not inflection points because \(f\) is undefined there. Thus, \(f^{\prime\prime}\) has \(0\) zeros. 2. In panel b), the graph of \(g\) has one local minimum on the left branch and one local maximum on the right branch, so \(g^{\prime}\) has \(2\) zeros. The graph changes concavity once on each branch, so \(g^{\prime\prime}\) has \(2\) zeros.

a) \(f^{\prime}\) has \(1\) zero, and \(f^{\prime\prime}\) has \(0\) zeros. b) \(g^{\prime}\) has \(2\) zeros, and \(g^{\prime\prime}\) has \(2\) zeros.

- Zeros of the first derivative occur where the graph has a horizontal tangent line. - Zeros of the second derivative can occur where the graph changes concavity. - Analyze each continuous branch separately; a vertical asymptote is not an extremum or an inflection point.

1. For \(h\), the graph has a local minimum at \(x=-1\) and a local maximum at \(x=1\). These are the two points with horizontal tangent lines, so \(h^{\prime}\) has \(2\) zeros. 2. The graph of \(h\) changes concavity at \(x=-\sqrt{3}\), \(x=0\), and \(x=\sqrt{3}\). Therefore, \(h^{\prime\prime}\) has \(3\) zeros. 3. For \(k\), the graph has one local maximum at \(x=2\). Therefore, \(k^{\prime}\) has \(1\) zero. 4. The right branch of \(k\) changes concavity once, at \(x=3\). The left branch has no concavity change, and the vertical asymptote is not an inflection point. Therefore, \(k^{\prime\prime}\) has \(1\) zero.

Graph \(h\): \(h^{\prime}\) has \(2\) zeros, and \(h^{\prime\prime}\) has \(3\) zeros. Graph \(k\): \(k^{\prime}\) has \(1\) zero, and \(k^{\prime\prime}\) has \(1\) zero.

- Differentiate the polynomial factor and the exponential factor separately. - Then group terms by powers of \(x\). - Compare the constant terms for consecutive indices.

1. Base case: For \(n=0\), the formula gives \(f^{(0)}(x)=x^2e^x\). 2. Inductive hypothesis: Assume \(f^{(k)}(x)=\left(x^2+2kx+k(k-1)\right)e^x\) for an arbitrary integer \(k\ge0\). 3. Using the product rule, \(f^{(k+1)}(x)=\left(2x+2k+x^2+2kx+k(k-1)\right)e^x\). 4. Combine like terms: \(f^{(k+1)}(x)=\left(x^2+2(k+1)x+k(k+1)\right)e^x\). 5. This is the formula for \(k+1\), so it holds for every integer \(n\ge0\).

For every integer \(n\ge0\), \(f^{(n)}(x)=\left(x^2+2nx+n(n-1)\right)e^x\).