Free math worksheets

- The quotient-rule numerator contains a subtraction. - The denominator in the derivative is the square of the original denominator. - Dividing the two separate derivatives is not the quotient rule.

1. The quotient rule uses the derivative of the numerator times the denominator minus the numerator times the derivative of the denominator. 2. The entire numerator is divided by the square of the original denominator. 3. Therefore, formula B is correct.

B) \(\frac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2}\)

- Write the quotient rule symbolically before substituting the data. - Keep the subtraction in the numerator, especially because one supplied derivative value is negative. - Square the original denominator value, not its derivative.

1. By the quotient rule, \(h'(1)=\frac{u'(1)v(1)-u(1)v'(1)}{[v(1)]^2}\). 2. Substitute the values: \(h'(1)=\frac{2(3)-6(-1)}{3^2}=\frac{12}{9}=\frac{4}{3}\).

\(h'(1)=\frac{4}{3}\)

- Use the quotient rule because both the numerator and denominator depend on \(x\). - Differentiate the numerator and denominator separately first. - Keep parentheses around complete numerator and denominator expressions when substituting. - Combine like terms in the resulting numerator.

1. Let \(u(x)=4x-1\) and \(v(x)=x^2+3\). Then \(u'(x)=4\) and \(v'(x)=2x\). 2. By the quotient rule, \(f'(x)=\frac{4(x^2+3)-(4x-1)(2x)}{(x^2+3)^2}\). 3. Simplify the numerator: \(4x^2+12-8x^2+2x=-4x^2+2x+12\). Therefore, \(f'(x)=\frac{-4x^2+2x+12}{(x^2+3)^2}\).

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

- Treat the parameter \(k\) as a constant. - Apply the quotient rule to the numerator and denominator. - Factor out the common factor \(k\) in the numerator.

1. Treat \(k\) as a constant. Let \(u(x)=kx\) and \(v(x)=x^2+1\). Then \(u'(x)=k\) and \(v'(x)=2x\). 2. By the quotient rule, \(f_k'(x)=\frac{k(x^2+1)-kx(2x)}{(x^2+1)^2}\). 3. Simplify and factor the numerator: \(k(x^2+1)-2kx^2=k(1-x^2)\). Therefore, \(f_k'(x)=\frac{k(1-x^2)}{(x^2+1)^2}\).

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

- Determine the logarithm's domain before differentiating. - For the Quotient Rule, identify the numerator and denominator and their derivatives. - Do not simplify the quotient-rule numerator until after part b.

1. The logarithm requires \(x>0\), which also keeps the denominator nonzero. Therefore, \(D_f=(0,\infty)\). 2. Applying the quotient rule gives \(f'(x)=\frac{(\frac1x)x-(\ln x)(1)}{x^2}\). 3. Simplifying gives \(f'(x)=\frac{1-\ln x}{x^2}\).

a) \(D_f=(0,\infty)\) b) \(f'(x)=\frac{(\frac1x)x-(\ln x)(1)}{x^2}\) c) \(f'(x)=\frac{1-\ln x}{x^2}\)

- Use the quotient rule. - Keep the subtraction and parentheses in the numerator. - Combine like exponential terms after expanding.

1. Let \(u(x)=e^x+1\) and \(v(x)=e^x-1\). Then \(u'(x)=e^x\) and \(v'(x)=e^x\). 2. By the quotient rule, \(h'(x)=\frac{e^x(e^x-1)-(e^x+1)e^x}{(e^x-1)^2}\). 3. The numerator simplifies to \(e^{2x}-e^x-e^{2x}-e^x=-2e^x\). Therefore, \(h'(x)=\frac{-2e^x}{(e^x-1)^2}\), for \(x\ne 0\).

\(h'(x)=\frac{-2e^x}{(e^x-1)^2}\), for \(x\ne 0\)

- The derivative of a constant denominator is zero. - Rewrite division by \(8\) as multiplication by \(\frac{1}{8}\). - Compare the number of algebraic steps in the two methods. - Notice which quotient-rule term vanishes.

1. Using the quotient rule with constant denominator \(8\), \(h'(x)=\frac{(3x^2-5)\cdot 8-(x^3-5x+2)\cdot 0}{8^2}\) \(=\frac{3x^2-5}{8}\). 2. Rewrite the function as \(h(x)=\frac{1}{8}(x^3-5x+2)\). Then \(h'(x)=\frac{1}{8}(3x^2-5)\). 3. The constant-multiple rule avoids squaring the denominator and then canceling the same constant factor.

1) \(h'(x)=\frac{3x^2-5}{8}\) 2) \(h(x)=\frac{1}{8}(x^3-5x+2)\), so \(h'(x)=\frac{1}{8}(3x^2-5)\). 3) The constant-multiple rule requires fewer algebraic steps.

- Divide each numerator term separately by the denominator. - A sum of powers is usually easier to differentiate than a quotient. - Track the negative exponent when differentiating \(x^{-2}\). - Simplify the quotient-rule result by factoring and canceling.

1. For a), divide term by term: \(g(x)=2x+1+x^{-2}\). Therefore, \(g'(x)=2-2x^{-3}=2-\frac{2}{x^3}\). 2. For b), use the quotient rule: \(g'(x)=\frac{(6x^2+2x)x^2-(2x^3+x^2+1)(2x)}{x^4}\) \(=\frac{2x^4-2x}{x^4}\) \(=2-\frac{2}{x^3}\). 3. Dividing term by term is more efficient because it reduces the problem to the power rule and avoids the longer quotient-rule simplification.

a) \(g'(x)=2-\frac{2}{x^3}\) b) The quotient rule gives the same derivative. Term-by-term division is more efficient.

- Identify the numerator and denominator functions. - Recall the derivatives of sine and a quadratic polynomial. - Apply the quotient rule. - Leave the denominator as a square.

1. Let \(u(x)=\sin(x)\) and \(v(x)=x^2+3\). Then \(u'(x)=\cos(x)\) and \(v'(x)=2x\). 2. By the quotient rule, \(f'(x)=\frac{\cos(x)(x^2+3)-\sin(x)(2x)}{(x^2+3)^2}\). Therefore, \(f'(x)=\frac{(x^2+3)\cos(x)-2x\sin(x)}{(x^2+3)^2}\).

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

- Use the quotient rule. - Differentiate the numerator and denominator separately first. - Keep the second product in parentheses because it is subtracted. - Leave the denominator as a square.

1. Let \(u(x)=0.5x^2+4x-1\) and \(v(x)=2x+3\). Then \(u'(x)=x+4\) and \(v'(x)=2\). 2. By the quotient rule, \(f'(x)=\frac{(x+4)(2x+3)-2(0.5x^2+4x-1)}{(2x+3)^2}\). 3. Simplify the numerator: \((2x^2+11x+12)-(x^2+8x-2)=x^2+3x+14\). Therefore, \(f'(x)=\frac{x^2+3x+14}{(2x+3)^2}\).

\(f'(x)=\frac{x^2+3x+14}{(2x+3)^2}\)

- Use the quotient rule. - Write the numerator function, denominator function, and their derivatives before substituting. - Keep parentheses around the second product in the numerator. - Factor the original denominator to identify excluded values.

1. Let \(u(x)=x^4-2\) and \(v(x)=x^2+x\). Then \(u'(x)=4x^3\) and \(v'(x)=2x+1\). 2. By the quotient rule, \(h'(x)=\frac{4x^3(x^2+x)-(x^4-2)(2x+1)}{(x^2+x)^2}\). 3. Expand and combine like terms in the numerator: \(4x^5+4x^4-(2x^5+x^4-4x-2)=2x^5+3x^4+4x+2\). Therefore, \(h'(x)=\frac{2x^5+3x^4+4x+2}{(x^2+x)^2}\). The derivative is defined for \(x\ne 0\) and \(x\ne -1\).

\(h'(x)=\frac{2x^5+3x^4+4x+2}{(x^2+x)^2}\), with \(x\ne 0\) and \(x\ne -1\)

- Use the quotient rule. - Differentiate the numerator and denominator separately. - Factor the numerator after applying the rule. - Cancel only factors, not individual terms.

1. Let \(u(x)=x+5\) and \(v(x)=x^3\). Then \(u'(x)=1\) and \(v'(x)=3x^2\). 2. By the quotient rule, \(f'(x)=\frac{x^3-3x^2(x+5)}{x^6}\). 3. Simplify and cancel a common factor of \(x^2\): \(f'(x)=\frac{-2x^3-15x^2}{x^6}=\frac{-2x-15}{x^4}\), where \(x\ne 0\).

\(f'(x)=\frac{-2x-15}{x^4}\), with \(x\ne 0\)

- Use the quotient rule. - Treat the parameter \(t\) as a constant when differentiating with respect to \(x\). - Combine like terms in the numerator before factoring. - Find the excluded input from the original denominator.

1. Let \(u(x)=tx^2\) and \(v(x)=x+t\). Treating \(t\) as a constant gives \(u'(x)=2tx\) and \(v'(x)=1\). 2. By the quotient rule, \(f_t'(x)=\frac{2tx(x+t)-tx^2}{(x+t)^2}\). 3. Simplify and factor the numerator: \(2tx^2+2t^2x-tx^2=tx^2+2t^2x=tx(x+2t)\). Thus, \(f_t'(x)=\frac{tx(x+2t)}{(x+t)^2}\), with \(x\ne -t\).

\(f_t'(x)=\frac{tx(x+2t)}{(x+t)^2}\), with \(x\ne -t\)

- Treat \(a\) as a constant while differentiating with respect to \(x\). - Apply the quotient rule before evaluating at \(x=0\). - Simplify the quotient-rule numerator carefully because several parameter terms cancel.

1. Using the quotient rule, \(g_a'(x)=\frac{-(ax+1)-a(a-x)}{(ax+1)^2}\). 2. The numerator simplifies to \(-(1+a^2)\), so \(g_a'(x)=-\frac{1+a^2}{(ax+1)^2}\). 3. At \(x=0\), the slope is \(g_a'(0)=-(1+a^2)\).

\(g_a'(0)=-(1+a^2)\)

- Keep the numerator order of the Quotient Rule consistent before using the table. - The denominator uses the square of the denominator function's value, not its derivative. - Substitute the four table entries only after writing the complete rule.

1. Use the Quotient Rule: \(q'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}\). 2. Substitute the table values at \(x=-1\): \(q'(-1)=\frac{(-2)(3)-6(4)}{3^2}=\frac{-30}{9}=-\frac{10}{3}\).

\(q'(-1)=-\frac{10}{3}\)

- A rational function is zero when its numerator is zero and its denominator is not. - Use the quotient rule to find the derivative. - A tangent line requires a point and a slope. - Use the intercepts as the triangle's base and the tangent intersection to find its height.

1. The numerator is zero at \(x=-2\) and \(x=2\), while the denominator is nonzero. The x-intercepts are \((-2,0)\) and \((2,0)\). 2. Differentiate using the quotient rule: \(f'(x)=\frac{2x(x^2+4)-(x^2-4)(2x)}{(x^2+4)^2}=\frac{16x}{(x^2+4)^2}\). 3. The slopes are \(f'(-2)=-\frac{1}{2}\) and \(f'(2)=\frac{1}{2}\). 4. The tangent lines are \(y=-\frac{1}{2}(x+2)=-\frac{1}{2}x-1\) and \(y=\frac{1}{2}(x-2)=\frac{1}{2}x-1\). 5. Set the line equations equal. This gives \(x=0\) and \(y=-1\), so \(S=(0,-1)\). 6. The base on the x-axis has length \(4\), and the height is \(1\). Therefore, the area is \(\frac{1}{2}\cdot 4\cdot 1=2\).

a) \((-2,0)\) and \((2,0)\) b) \(y=-\frac{1}{2}x-1\) and \(y=\frac{1}{2}x-1\) c) \(S=(0,-1)\); area \(=2\) square units

- Expand the squared denominator before differentiating so the quotient rule is the only new rule needed. - Keep the subtraction in the quotient-rule numerator grouped until both products are complete. - Factor the final expression to expose any cancellation.

1. Expand the denominator before differentiating: \((2x-1)^2=4x^2-4x+1\). 2. Let \(u(x)=x^2+1\) and \(v(x)=4x^2-4x+1\). Then \(u'(x)=2x\) and \(v'(x)=8x-4\). 3. Apply the quotient rule: \(f'(x)=\frac{2x(4x^2-4x+1)-(x^2+1)(8x-4)}{(4x^2-4x+1)^2}\). 4. Simplifying and factoring gives \(f'(x)=\frac{-2(x+2)}{(2x-1)^3}\).

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

- Identify the numerator and denominator functions first. - Differentiate those two functions before substituting into the quotient rule. - Track the negative sign before the second product in the quotient-rule numerator. - Usually the denominator can remain as a square.

1. For a), \(f'(x)=\frac{0(x^2+1)-5(2x)}{(x^2+1)^2}=\frac{-10x}{(x^2+1)^2}\). 2. For b), \(f'(x)=\frac{1(x-3)-x(1)}{(x-3)^2}=\frac{-3}{(x-3)^2}\). 3. For c), \(f'(x)=\frac{2x(x)-(x^2+4)(1)}{x^2}=\frac{x^2-4}{x^2}\). 4. For d), \(f'(x)=\frac{2(x^2+2)-(2x-1)(2x)}{(x^2+2)^2}\) \(=\frac{-2x^2+2x+4}{(x^2+2)^2}\).

a) \(f'(x)=\frac{-10x}{(x^2+1)^2}\) b) \(f'(x)=\frac{-3}{(x-3)^2}\) c) \(f'(x)=\frac{x^2-4}{x^2}\) d) \(f'(x)=\frac{-2x^2+2x+4}{(x^2+2)^2}\)

- Differentiate the two terms separately. - Rewrite the first denominator using a negative exponent. - Use the quotient rule for the second term. - Expand and combine the quotient-rule numerator carefully.

1. Rewrite the first term as \(3x^{-4}\). Its derivative is \(-12x^{-5}=-\frac{12}{x^5}\). 2. For the second term, let \(u(x)=x^2\) and \(v(x)=2x-1\). By the quotient rule, \(\left(\frac{x^2}{2x-1}\right)'=\frac{2x(2x-1)-x^2\cdot 2}{(2x-1)^2}\). 3. Simplify the numerator: \(2x(2x-1)-2x^2=2x^2-2x=2x(x-1)\). Therefore, \(f'(x)=-\frac{12}{x^5}+\frac{2x(x-1)}{(2x-1)^2}\).

\(f'(x)=-\frac{12}{x^5}+\frac{2x(x-1)}{(2x-1)^2}\)

- Compare each calculation with the quotient rule. - Check the structure of the denominator. - Check the operation between the two products in the numerator. - Distribute a negative sign carefully when simplifying.

1. For \(f\), the denominator in the quotient rule must be squared. The correct derivative is \(f'(x)=\frac{2x(2x-3)-2x^2}{(2x-3)^2}\) \(=\frac{2x^2-6x}{(2x-3)^2}\). 2. For \(g\), the quotient-rule numerator requires subtraction, not addition. The correct derivative is \(g'(x)=\frac{3(x^2+5)-(3x-1)(2x)}{(x^2+5)^2}\) \(=\frac{-3x^2+2x+15}{(x^2+5)^2}\).

1) Error: The denominator was not squared. \(f'(x)=\frac{2x^2-6x}{(2x-3)^2}\) 2) Error: The two products in the numerator were added instead of subtracted. \(g'(x)=\frac{-3x^2+2x+15}{(x^2+5)^2}\)

- A horizontal tangent has slope zero. - A rational expression is zero when its numerator is zero and its denominator is nonzero. - Check that every solution is in the domain of the original function.

1. Let \(u(x)=x^2-8\) and \(v(x)=x-3\). Then \(u'(x)=2x\) and \(v'(x)=1\). 2. By the quotient rule, \(f'(x)=\frac{2x(x-3)-(x^2-8)}{(x-3)^2}\) \(=\frac{x^2-6x+8}{(x-3)^2}\). 3. A horizontal tangent occurs where \(f'(x)=0\). Since the denominator is nonzero in the domain, solve \(x^2-6x+8=0\). Factoring gives \((x-2)(x-4)=0\), so \(x=2\) or \(x=4\). Both values are in the domain. 4. Evaluate the function at those inputs: \(f(2)=4\) and \(f(4)=8\). Therefore, the horizontal tangent points are \((2,4)\) and \((4,8)\).

a) \(f'(x)=\frac{x^2-6x+8}{(x-3)^2}\) b) The horizontal tangent points are \((2,4)\) and \((4,8)\).

- The derivative gives the tangent slope. - Use the quotient rule to find the derivative. - Set the derivative equal to the given slope. - Substitute each resulting \(x\)-value into the original function to find the point.

1. By the quotient rule, \(f'(x)=\frac{2x(x+2)-x^2}{(x+2)^2}=\frac{x^2+4x}{(x+2)^2}\). 2. Set the derivative equal to the desired slope: \(\frac{x^2+4x}{(x+2)^2}=\frac{3}{4}\). 3. Cross-multiply and simplify: \(4(x^2+4x)=3(x^2+4x+4)\), so \(x^2+4x-12=0\). Factoring gives \(x=2\) or \(x=-6\). 4. Find the corresponding function values: \(f(2)=1\) and \(f(-6)=-9\). Therefore, the points are \((2,1)\) and \((-6,-9)\).

The points are \((2,1)\) and \((-6,-9)\).

- The derivative gives the slope of the graph. - Apply the quotient rule. - Exclude any value that makes the original denominator zero. - Substitute each solution into the original function to find the full coordinates.

1. By the quotient rule, \(f'(x)=\frac{2(x-2)-(2x+2)}{(x-2)^2}=\frac{-6}{(x-2)^2}\). 2. Set the derivative equal to the given slope: \(\frac{-6}{(x-2)^2}=-1.5\). Thus, \((x-2)^2=4\), so \(x=4\) or \(x=0\). 3. Evaluate the original function: \(f(4)=5\) and \(f(0)=-1\). Therefore, the points are \((4,5)\) and \((0,-1)\).

The points are \((4,5)\) and \((0,-1)\).

- A tangent line requires a point and a slope. - The function value gives the point, and the derivative gives the slope. - Use the quotient rule to find the derivative. - Apply point-slope form.

1. Find the point on the graph: \(f(1)=\frac{3}{2}=1.5\). 2. By the quotient rule, \(f'(x)=\frac{(x^2+1)-(x+2)(2x)}{(x^2+1)^2}\) \(=\frac{-x^2-4x+1}{(x^2+1)^2}\). 3. The tangent slope is \(f'(1)=\frac{-1-4+1}{4}=-1\). 4. Use point-slope form through \((1,1.5)\): \(y-1.5=-(x-1)\), so \(y=-x+2.5\).

The tangent line is \(y=-x+2.5\).

- Apply the quotient rule in part a. - In part b, either combine the terms or differentiate the negative powers directly. - Track the subtraction in the quotient-rule numerator. - Compare the signs of the final numerators carefully.

1. For a), apply the quotient rule: \(f'(x)=\frac{2x(0.4x^2-1)-(x^2+2)(0.8x)}{(0.4x^2-1)^2}\). The numerator simplifies to \(0.8x^3-2x-0.8x^3-1.6x=-3.6x\). Therefore, \(f'(x)=\frac{-3.6x}{(0.4x^2-1)^2}=g(x)\). 2. For b), use the power rule: \(f'(x)=-8x^{-3}-x^{-2}\) \(=\frac{-8-x}{x^3}\). This is the negative of the proposed numerator, so \(g\) is not the derivative of \(f\).

a) Yes. \(g(x)=f'(x)\). b) No. The correct derivative is \(f'(x)=\frac{-x-8}{x^3}\).

- Keep the numerator and denominator roles separate when applying the Quotient Rule. - Simplify the derivative only after the Quotient Rule has been set up correctly. - For a rational derivative, determine when its numerator can be zero while its denominator remains defined.

1. Apply the Quotient Rule to the original quotient: \(f'(x)=\frac{2x(x-1)-x^2}{(x-1)^2}\). 2. Simplify the numerator: \(2x(x-1)-x^2=x^2-2x=x(x-2)\). Therefore, \(f'(x)=\frac{x(x-2)}{(x-1)^2}\). 3. A horizontal tangent requires \(f'(x)=0\). Since \(x=1\) is excluded from the domain and the denominator is nonzero elsewhere, set the numerator equal to zero: \(x(x-2)=0\). Thus, \(x=0\) or \(x=2\). 4. Jordan's claim is complete. Maya omitted \(x=0\).

\(f'(x)=\frac{x(x-2)}{(x-1)^2}\). The horizontal tangents occur at \(x=0\) and \(x=2\), so Jordan's claim is complete and Maya's is not.

- What derivative condition defines a horizontal tangent? - Distinguish between zeros of a function and zeros of its derivative. - Use the quotient rule to find the slope function. - When is a rational expression equal to zero?

1. Apply the quotient rule: \(g'(x) = \frac{2x(x^2 + 1) - 2x(x^2 - 4)}{(x^2 + 1)^2}\). 2. Simplify: \(g'(x) = \frac{10x}{(x^2 + 1)^2}\). 3. A horizontal tangent occurs where \(g'(x) = 0\). Since the denominator is always positive, solve \(10x = 0\), which gives \(x = 0\). 4. The student confused zeros of the function with zeros of its derivative. At \(x = 2\), \(g'(2) = \frac{4}{5}\), and at \(x = -2\), \(g'(-2) = -\frac{4}{5}\), so neither tangent is horizontal.

The claim is false. The only horizontal tangent occurs at \(x = 0\). The values \(x = \pm 2\) are zeros of \(g\), not zeros of \(g'\).

- Apply the quotient rule. - Determine which factors can equal zero. - Check that the derivative changes sign at the solutions.

1. The quotient rule gives \(f_a'(x)=a\frac{1-x^2}{(x^2+x+1)^2}\). 2. The denominator is positive for every real \(x\), and \(a\ne0\). Therefore, \(f_a'(x)=0\) exactly when \(1-x^2=0\). 3. The solutions are \(x=-1\) and \(x=1\). The numerator changes sign at both values, so both are local-extremum x-coordinates for every \(a\).

\(x=-1\) and \(x=1\)

- Identify the numerator and denominator functions and their derivatives before comparing the student's expression with the Quotient Rule. - Check both products in the derivative numerator; each must contain the appropriate derivative factor. - After correcting the derivative, use the fact that the denominator is always positive to locate a zero of the derivative.

1. The denominator is \(x^2+1\), whose derivative is \(2x\). In the Quotient Rule numerator, the second product must therefore include this factor. The student omitted it. 2. Apply the Quotient Rule correctly: \(f'(x)=\frac{2x(x^2+1)-(x^2-3)(2x)}{(x^2+1)^2}\). 3. Simplify the numerator: \(2x[(x^2+1)-(x^2-3)]=2x(4)=8x\). Thus, \(f'(x)=\frac{8x}{(x^2+1)^2}\). 4. Since \((x^2+1)^2>0\) for every real \(x\), the derivative is zero exactly when \(8x=0\). Hence the graph has a horizontal tangent at \(x=0\).

The student omitted the derivative factor \(2x\) from the second product in the Quotient Rule numerator. The correct derivative is \(f'(x)=\frac{8x}{(x^2+1)^2}\), so the horizontal tangent occurs at \(x=0\).

- 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 \(x=0\).

- Differentiate the two terms separately. - Use the quotient rule for the exponential fraction. - Factor out the common exponential factor in the quotient-rule numerator.

1. Differentiate the polynomial term: \(\frac{d}{dx}(4x^2)=8x\). 2. For the quotient, let \(u(x)=e^x\) and \(v(x)=x+3\). Then \(u'(x)=e^x\) and \(v'(x)=1\). 3. By the quotient rule, \(\frac{d}{dx}\left(\frac{e^x}{x+3}\right)=\frac{e^x(x+3)-e^x}{(x+3)^2}\) \(=\frac{(x+2)e^x}{(x+3)^2}\). Therefore, \(f'(x)=8x+\frac{(x+2)e^x}{(x+3)^2}\).

\(f'(x)=8x+\frac{(x+2)e^x}{(x+3)^2}\)

- Use the quotient rule for the overall expression. - The numerator requires the product rule. - Expand the quotient-rule numerator carefully. - Look for terms that cancel.

1. Let \(u(t)=t\sin(t)\) and \(v(t)=t-1\). 2. Use the product rule on the numerator: \(u'(t)=\sin(t)+t\cos(t)\). Also, \(v'(t)=1\). 3. By the quotient rule, \(f'(t)=\frac{[\sin(t)+t\cos(t)](t-1)-t\sin(t)}{(t-1)^2}\). 4. Expand and combine terms in the numerator: \(t\sin(t)-\sin(t)+t^2\cos(t)-t\cos(t)-t\sin(t)\) \(=(t^2-t)\cos(t)-\sin(t)\). Therefore, \(f'(t)=\frac{(t^2-t)\cos(t)-\sin(t)}{(t-1)^2}\).

\(f'(t)=\frac{(t^2-t)\cos(t)-\sin(t)}{(t-1)^2}\)

- 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 \(x=1\).

- Apply the quotient rule in each part. - Use the basic derivative of \(e^t\) in part b. - Factor the completed numerator before deciding that the expression is simplified.

1. For part a, the quotient rule gives \(f'(x)=\frac{2x(3x+2)-3(x^2-9)}{(3x+2)^2}=\frac{3x^2+4x+27}{(3x+2)^2}\). 2. For part b, the quotient rule gives \(g'(t)=\frac{e^t(t^2+1)-e^t(2t)}{(t^2+1)^2}\). 3. Factor the numerator: \(g'(t)=\frac{e^t(t^2-2t+1)}{(t^2+1)^2}=\frac{e^t(t-1)^2}{(t^2+1)^2}\).

a) \(f'(x)=\frac{3x^2+4x+27}{(3x+2)^2}\) b) \(g'(t)=\frac{e^t(t-1)^2}{(t^2+1)^2}\)

- Differentiate the terms of a sum separately. - The denominator in part b is a product, so differentiate it with the product rule. - Factor common terms before canceling. - Recall the product rule for \(x^2e^x\).

1. For a), differentiate each term: \(f'(x)=x-\frac{4}{(x-5)^2}\). 2. For b), let \(u(x)=3x+2\) and \(v(x)=x^2e^x\). By the product rule, \(v'(x)=2xe^x+x^2e^x=xe^x(x+2)\). 3. Apply the quotient rule: \(f'(x)=\frac{3x^2e^x-(3x+2)xe^x(x+2)}{x^4e^{2x}}\). 4. Factor and cancel \(xe^x\): \(f'(x)=\frac{3x-(3x+2)(x+2)}{x^3e^x}\) \(=\frac{-3x^2-5x-4}{x^3e^x}\).

a) \(f'(x)=x-\frac{4}{(x-5)^2}\) b) \(f'(x)=\frac{-3x^2-5x-4}{x^3e^x}\)

- Treat \(k\) as a constant. - Rewrite \(x^{-2}\) carefully for the product-rule method. - Simplify the quotient-rule result by canceling a factor of \(x\). - Use a common denominator to compare the two forms.

1. Using the quotient rule, \(f_k'(x)=\frac{-kx^2-(1-kx)(2x)}{x^4}\) \(=\frac{kx^2-2x}{x^4}=\frac{kx-2}{x^3}\). 2. Rewrite the function as \(f_k(x)=(1-kx)x^{-2}\). By the product rule, \(f_k'(x)=-kx^{-2}+(1-kx)(-2x^{-3})\) \(=kx^{-2}-2x^{-3}\). 3. Combining the terms gives \(\frac{k}{x^2}-\frac{2}{x^3}=\frac{kx-2}{x^3}\), so the methods agree.

\(f_k'(x)=\frac{kx-2}{x^3}\), for \(x\ne 0\)

- Expand the squared denominator so differentiating it does not require a later rule. - In the quotient rule, identify the numerator and denominator functions before substituting their derivatives. - Determine the sign only after simplifying the derivative. - Interpret the limiting illuminance using the model's units.

1. \(I(0)=\frac{800}{4}=200\), so the modeled illuminance at \(d=0\) is \(200\,\text{lx}\). 2. Expand \((d+2)^2=d^2+4d+4\). With \(u(d)=800\) and \(v(d)=d^2+4d+4\), \(u'(d)=0\) and \(v'(d)=2d+4\). 3. The quotient-rule expression is \(I'(d)=\frac{0\cdot(d^2+4d+4)-800(2d+4)}{(d^2+4d+4)^2}\). 4. Simplifying gives \(I'(d)=-\frac{1600}{(d+2)^3}\). For \(d\ge0\), the denominator is positive, so \(I'(d)<0\). 5. As \(d\to\infty\), the denominator grows without bound, so \(I(d)\to0\). In the model, illuminance approaches \(0\,\text{lx}\) at very large distances.

a) \(200\,\text{lx}\); this is the modeled illuminance at \(d=0\) b) \(I'(d)=\frac{0\cdot(d^2+4d+4)-800(2d+4)}{(d^2+4d+4)^2}=-\frac{1600}{(d+2)^3}<0\) for \(d\ge0\) c) \(\lim_{d\to\infty}I(d)=0\); the modeled illuminance approaches \(0\,\text{lx}\)

- Substitute \(t=0\). - Apply the quotient rule carefully. - The denominator of the derivative is always positive, so analyze the numerator. - Compare the degrees of the numerator and denominator for the limit.

1. \(c(0)=0\,\text{mg/L}\). 2. Apply the quotient rule: \(c'(t)=\frac{10(t^2+4)-10t(2t)}{(t^2+4)^2}=\frac{40-10t^2}{(t^2+4)^2}\). 3. The denominator is always positive. The numerator is positive for \(0\le t<2\), zero at \(t=2\), and negative for \(t>2\). Therefore, the concentration increases on \([0,2]\) and decreases for \(t>2\). In the model, the medication level rises until \(2\) hours after administration and then falls. 4. Since the denominator has higher degree than the numerator, \(\lim_{t\to\infty}c(t)=0\). The modeled concentration approaches \(0\,\text{mg/L}\).

a) \(0\,\text{mg/L}\). b) \(c'(t)=\frac{40-10t^2}{(t^2+4)^2}\). c) Increasing on \([0,2]\); decreasing for \(t>2\). The concentration rises and then falls after reaching its peak at \(t=2\). d) \(0\,\text{mg/L}\).

- 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 \(x\ne\frac{3}{2}\).

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

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

- 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.

- 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.

- 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: \(g'(x)=\frac{4(x^2+3)-4x^2}{(x^2+3)\sqrt{x^2+3}}=\frac{12}{(x^2+3)\sqrt{x^2+3}}\).

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

- 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 \(D_f\)

- Rewrite the function before differentiating. - Write the tangent line at a general x-coordinate \(x_0\). - Use the fact that the origin lies on the tangent line. - Substitute the resulting x-coordinate into the original function.

1. Rewrite the function as \(f_k(x)=1+\frac{k}{x}\), so \(f_k'(x)=-\frac{k}{x^2}\). 2. Let the point of tangency have x-coordinate \(x_0\). The tangent line is \(y=f_k'(x_0)(x-x_0)+f_k(x_0)\). 3. Since the line passes through \((0, 0)\), \(0=-\frac{k}{x_0^2}(0-x_0)+1+\frac{k}{x_0}\). 4. Simplify: \(0=1+\frac{2k}{x_0}\), so \(x_0=-2k\). Since \(k>0\), this is a valid domain value. 5. The y-coordinate is \(f_k(-2k)=\frac{-2k+k}{-2k}=\frac{1}{2}\). Therefore, \(B=\left(-2k, \frac{1}{2}\right)\).

\(B=\left(-2k, \frac{1}{2}\right)\)

- Use a logarithm property to determine the derivative of the numerator without a later Chain Rule. - Identify the numerator and denominator functions before applying the Quotient Rule. - Keep the quotient-rule numerator intact until part a is complete.

1. Since \(x>0\), \(\ln(5x)=\ln5+\ln x\), so the numerator derivative is \(\frac1x\). 2. With numerator \(u(x)=\ln(5x)\) and denominator \(v(x)=x^2\), the quotient rule gives \(f'(x)=\frac{(\frac1x)x^2-\ln(5x)(2x)}{x^4}\). 3. Simplifying gives \(f'(x)=\frac{1-2\ln(5x)}{x^3}\).

a) \(f'(x)=\frac{(\frac1x)x^2-\ln(5x)(2x)}{x^4}\) b) \(f'(x)=\frac{1-2\ln(5x)}{x^3}\)

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

- Use the quotient rule. - Determine what happens to \(e^x\) as \(x\to-\infty\). - For \(x\to\infty\), divide the numerator and denominator by \(e^x\). - A finite end-behavior limit gives a horizontal asymptote.

1. Apply the quotient rule: \(g'(x)=\frac{e^x(e^x+2)-e^x(e^x)}{(e^x+2)^2}=\frac{2e^x}{(e^x+2)^2}\). 2. As \(x\to-\infty\), \(e^x\to0\), so \(g(x)\to0\). Thus, \(y=0\) is a horizontal asymptote. 3. As \(x\to\infty\), divide numerator and denominator by \(e^x\): \(g(x)=\frac{1}{1+2e^{-x}}\to1\). Thus, \(y=1\) is a horizontal asymptote.

a) \(g'(x)=\frac{2e^x}{(e^x+2)^2}\) b) \(\lim_{x\to-\infty}g(x)=0\) and \(\lim_{x\to\infty}g(x)=1\); horizontal asymptotes: \(y=0\) and \(y=1\)

- Use \(\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}\) for the first derivative of a quotient. - Differentiate the numerator and denominator separately only for the comparison in part b). - Determine whether the two resulting expressions are equal for every allowed input.

1. By the quotient rule, \(k'(x)=\frac{(6x^5)(x^2)-(x^6)(2x)}{(x^2)^2}\). 2. Simplify: \(k'(x)=\frac{6x^7-2x^7}{x^4}=4x^3\). 3. Differentiating numerator and denominator separately and dividing gives \(\frac{6x^5}{2x}=3x^4\). 4. The expressions \(4x^3\) and \(3x^4\) are equal only when \(x=\frac{4}{3}\) in the domain \(x>0\). They are not identical, so separate differentiation is not a valid quotient rule.

a) \(k'(x)=4x^3\) b) \(3x^4\) c) No. The expressions agree only at \(x=\frac{4}{3}\), not for every \(x>0\).

- Identify the numerator and denominator in each part. - Differentiate those components separately. - Track the subtraction in the quotient-rule numerator. - Factor a common \(x\) in part b to simplify.

1. For a), \(f'(x)=\frac{4x(3x+4)-3(2x^2-5)}{(3x+4)^2}\) \(=\frac{6x^2+16x+15}{(3x+4)^2}\). 2. For b), \(g'(x)=\frac{-\sin(x)x^2-\cos(x)(2x)}{x^4}\) \(=\frac{-x\sin(x)-2\cos(x)}{x^3}\).

a) \(f'(x)=\frac{6x^2+16x+15}{(3x+4)^2}\), for \(x\ne-\frac{4}{3}\) b) \(g'(x)=\frac{-x\sin(x)-2\cos(x)}{x^3}\), for \(x\ne 0\)

- In the quotient rule, the derivative of a constant numerator is zero. - Constant factors remain when differentiating. - Differentiate the denominator in part b first.

1. For a), let the numerator be \(u(x)=1\). Then \(u'(x)=0\). By the quotient rule, \(h'(x)=\frac{0\cdot v(x)-1\cdot v'(x)}{(v(x))^2}\) \(=-\frac{v'(x)}{(v(x))^2}\). 2. For b), let \(v(x)=x^2+\sin(x)\). Then \(v'(x)=2x+\cos(x)\). Thus, \(f'(x)=-5\frac{2x+\cos(x)}{(x^2+\sin(x))^2}\) \(=-\frac{10x+5\cos(x)}{(x^2+\sin(x))^2}\).

a) \(h'(x)=-\frac{v'(x)}{(v(x))^2}\), wherever \(v(x)\ne 0\) b) \(f'(x)=-\frac{10x+5\cos(x)}{(x^2+\sin(x))^2}\), wherever \(x^2+\sin(x)\ne 0\)

- Treat the product in the denominator as one function first. - Apply the quotient rule to the full fraction. - Use the product rule when differentiating the denominator. - Distribute the negative sign across both terms.

1. Treat \(v(x)w(x)\) as the denominator. By the quotient rule, \(f'(x)=\frac{u'(x)v(x)w(x)-u(x)[v(x)w(x)]'}{[v(x)w(x)]^2}\). 2. By the product rule, \([v(x)w(x)]'=v'(x)w(x)+v(x)w'(x)\). 3. Substitute and distribute the negative sign: \(f'(x)=\frac{u'(x)v(x)w(x)-u(x)v'(x)w(x)-u(x)v(x)w'(x)}{v(x)^2w(x)^2}\).

\(f'(x)=\frac{u'(x)v(x)w(x)-u(x)v'(x)w(x)-u(x)v(x)w'(x)}{v(x)^2w(x)^2}\)

- The function contains both a product and a quotient. - Differentiate the numerator and denominator separately first. - Factor the common exponential term before simplifying.

1. Let \(u(x)=x\sin(x)\) and \(v(x)=e^x\). 2. By the product rule, \(u'(x)=\sin(x)+x\cos(x)\), and \(v'(x)=e^x\). 3. Apply the quotient rule: \(h'(x)=\frac{[\sin(x)+x\cos(x)]e^x-x\sin(x)e^x}{e^{2x}}\). 4. Factor and cancel \(e^x\): \(h'(x)=\frac{\sin(x)+x\cos(x)-x\sin(x)}{e^x}\) \(=\frac{(1-x)\sin(x)+x\cos(x)}{e^x}\).

\(h'(x)=\frac{(1-x)\sin(x)+x\cos(x)}{e^x}\)

- Use polynomial division to rewrite the rational expression. - Differentiate the reciprocal term using a negative exponent. - Keep the subtraction in the quotient-rule numerator. - Use a common denominator to compare the two forms.

1. Polynomial division gives \(f(x)=x-2+\frac{1}{x-2}\). Therefore, \(f'(x)=1-\frac{1}{(x-2)^2}\). 2. By the quotient rule, \(f'(x)=\frac{(2x-4)(x-2)-(x^2-4x+5)}{(x-2)^2}\) \(=\frac{x^2-4x+3}{(x-2)^2}\). 3. Put the first result over a common denominator: \(1-\frac{1}{(x-2)^2}=\frac{(x-2)^2-1}{(x-2)^2}\) \(=\frac{x^2-4x+3}{(x-2)^2}\). Thus, the methods agree.

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

- Apply the quotient rule to \(\frac{1}{x^n}\). - The derivative of the constant numerator is zero. - Use exponent rules to simplify the quotient. - Treat the factor \(3\) as a constant.

1. For \(f(x)=\frac{1}{x^n}\), let \(u(x)=1\) and \(v(x)=x^n\). Then \(u'(x)=0\) and \(v'(x)=nx^{n-1}\). 2. By the quotient rule, \(f'(x)=\frac{0\cdot x^n-nx^{n-1}}{x^{2n}}\) \(=-n x^{n-1-2n}=-n x^{-n-1}\). 3. Apply the rule to \(g(x)=3x^{-5}\): \(g'(x)=3\cdot(-5)x^{-6}=-\frac{15}{x^6}\).

\(\frac{d}{dx}x^{-n}=-n x^{-n-1}\). For \(g(x)=\frac{3}{x^5}\), \(g'(x)=-\frac{15}{x^6}\).

- Divide each numerator term by \(x^4\) to obtain negative powers. - Track the subtraction in the quotient-rule numerator. - Rewrite both derivative forms with a common denominator to compare them. - Simplify the quotient-rule result by factoring powers of \(x\).

1. By the quotient rule, \(h'(x)=\frac{2x(x^4)-(x^2-5)(4x^3)}{x^8}\) \(=\frac{-2x^5+20x^3}{x^8}=\frac{-2x^2+20}{x^5}\). 2. Rewrite the function as \(h(x)=x^{-2}-5x^{-4}\). Then \(h'(x)=-2x^{-3}+20x^{-5}\). 3. Put the second result over the common denominator \(x^5\): \(-2x^{-3}+20x^{-5}=\frac{-2x^2+20}{x^5}\). Thus, the methods agree.

\(h'(x)=\frac{-2x^2+20}{x^5}=-2x^{-3}+20x^{-5}\)

- Rewrite the square root as a fractional power if helpful. - Track the signs when differentiating the denominator. - Keep parentheses around each quotient-rule component. - Clear the small fraction in the numerator using a common factor.

1. Let \(u(x)=\sqrt{x}\) and \(v(x)=\cos(x)-x\). Then \(u'(x)=\frac{1}{2\sqrt{x}}\) and \(v'(x)=-\sin(x)-1\). 2. By the quotient rule, \(f'(x)=\frac{\frac{1}{2\sqrt{x}}(\cos(x)-x)-\sqrt{x}(-\sin(x)-1)}{(\cos(x)-x)^2}\). 3. Multiply the numerator by a common factor of \(2\sqrt{x}\) and simplify: \(f'(x)=\frac{\cos(x)+x+2x\sin(x)}{2\sqrt{x}(\cos(x)-x)^2}\). The derivative is defined when \(x>0\) and \(\cos(x)\ne x\).

\(f'(x)=\frac{\cos(x)+x+2x\sin(x)}{2\sqrt{x}(\cos(x)-x)^2}\), with domain \(\{x\in\mathbb{R}:x>0\text{ and }\cos(x)\ne x\}\)

- Write the quotient rule before substituting. - Compare the degrees of the two products in the derivative numerator. - Check whether the leading terms cancel after subtraction. - Base your explanation on the simplified numerator.

1. Let \(u(x)=2x^3-5\) and \(v(x)=x^2+4\). Then \(u'(x)=6x^2\) and \(v'(x)=2x\). By the quotient rule, \(f'(x)=\frac{6x^2(x^2+4)-2x(2x^3-5)}{(x^2+4)^2}\) \(=\frac{6x^4+24x^2-4x^4+10x}{(x^2+4)^2}\) \(=\frac{2x^4+24x^2+10x}{(x^2+4)^2}\). 2. The numerator of \(f\) has degree \(3\). The simplified numerator of \(f'\) has degree \(4\). 3. In the quotient-rule numerator, both \(u'v\) and \(uv'\) have degree \(4\). Their leading terms are \(6x^4\) and \(4x^4\), so subtraction leaves the nonzero leading term \(2x^4\). Therefore, the numerator degree is \(4\), not less than \(3\).

1. \(f'(x)=\frac{2x^4+24x^2+10x}{(x^2+4)^2}\) 2. Original numerator degree: \(3\); derivative numerator degree: \(4\). 3. The two degree- \(4\) leading terms in the quotient-rule numerator do not cancel, so the simplified numerator still has degree \(4\).

- Use the quotient rule. - Differentiate sine and cosine carefully. - Look for the identity \(\sin^2x+\cos^2x=1\) in the numerator. - Cancel a common factor only after factoring the numerator.

1. Let \(u(x)=\sin x\) and \(v(x)=1+\cos x\). Then \(u'(x)=\cos x\) and \(v'(x)=-\sin x\). 2. Apply the quotient rule: \(f'(x)=\frac{\cos x(1+\cos x)-\sin x(-\sin x)}{(1+\cos x)^2}\). 3. Simplify the numerator using \(\sin^2x+\cos^2x=1\): \(\cos x+\cos^2x+\sin^2x=1+\cos x\). Therefore, \(f'(x)=\frac{1+\cos x}{(1+\cos x)^2}=\frac{1}{1+\cos x}\). The original function and its derivative are defined when \(x\ne (2k+1)\pi\), where \(k\in\mathbb{Z}\).

\(f'(x)=\frac{1}{1+\cos x}\), for \(x\ne (2k+1)\pi\), \(k\in\mathbb{Z}\)

- Apply the quotient rule before simplifying. - Use the given derivative for both occurrences of \(\tan x\). - Factor the common expression in the numerator instead of expanding it. - Check both restrictions in the original function.

1. Let \(u(x)=\tan x\) and \(v(x)=1+\tan x\). Then \(u'(x)=v'(x)=1+\tan^2x\). 2. Apply the quotient rule: \(f'(x)=\frac{(1+\tan^2x)(1+\tan x)-\tan x(1+\tan^2x)}{(1+\tan x)^2}\). 3. Factor the common expression in the numerator: \(f'(x)=\frac{(1+\tan^2x)[(1+\tan x)-\tan x]}{(1+\tan x)^2}\). Therefore, \(f'(x)=\frac{1+\tan^2x}{(1+\tan x)^2}\). The domain excludes values where \(\tan x\) is undefined and values where \(\tan x=-1\).

\(f'(x)=\frac{1+\tan^2x}{(1+\tan x)^2}\), for \(x\ne \frac{\pi}{2}+k\pi\) and \(x\ne -\frac{\pi}{4}+k\pi\), \(k\in\mathbb{Z}\)

- 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.

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

- 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 \(x>-\frac{2}{3}\)

- Rewrite the line in slope-intercept form. - Parallel lines have equal slopes. - The derivative gives the slope of the tangent line. - Which derivative rule applies to a quotient?

1. Rewrite the given line as \(y=-6x+4\), so its slope is \(-6\). 2. Differentiate using the quotient rule: \(f'(x)=\frac{4(x-1)-(4x+2)}{(x-1)^2}=\frac{-6}{(x-1)^2}\). 3. Parallel tangent lines must have slope \(-6\), so solve \(\frac{-6}{(x-1)^2}=-6\). 4. This gives \((x-1)^2=1\), so \(x-1=\pm1\). Therefore, \(x=0\) or \(x=2\). Both values are in the domain of \(f\).

\(x=0\) and \(x=2\)

- 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: \(f_k'(x)=\frac{k(x^2+1)-kx^2}{(x^2+1)\sqrt{x^2+1}}=\frac{k}{(x^2+1)^{\frac{3}{2}}}\).

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

- Identify the numerator and denominator before applying the quotient rule. - The denominator of the derivative is always positive. - Use a sign change to determine which critical point is the local maximum. - Eliminate \(k\) from the maximum coordinates.

1. The quotient rule gives \(g_k'(x)=\frac{(x^2+k)-x(2x)}{(x^2+k)^2}=\frac{k-x^2}{(x^2+k)^2}\). 2. The denominator is positive, so critical points satisfy \(k-x^2=0\), giving \(x=\pm\sqrt{k}\). The derivative changes from positive to negative at \(x=\sqrt{k}\), so the local maximum is \(\left(\sqrt{k}, \frac{1}{2\sqrt{k}}\right)\). 3. At the maximum, \(x=\sqrt{k}>0\), so \(k=x^2\). Substituting into the y-coordinate gives \(y=\frac{1}{2x}\). Thus the locus is \(y=\frac{1}{2x}\) for \(x>0\).

1. \(g_k'(x)=\frac{k-x^2}{(x^2+k)^2}\) 2. \(\left(\sqrt{k}, \frac{1}{2\sqrt{k}}\right)\) 3. \(y=\frac{1}{2x}\) for \(x>0\)

- Complete the Quotient Rule setup before using the tangent geometry. - Write the tangent at a general input \(x_0\) using the derivative you obtained. - Find the two axis intercepts separately, then apply the midpoint formula.

1. With numerator \(4\) and denominator \(x\), the quotient rule gives \(f'(x)=\frac{0\cdot x-4\cdot1}{x^2}=-\frac4{x^2}\). 2. At \(P(x_0,\frac4{x_0})\), the tangent is \(y=-\frac4{x_0^2}(x-x_0)+\frac4{x_0}=-\frac4{x_0^2}x+\frac8{x_0}\). 3. Setting \(x=0\) gives \(S_y=(0,\frac8{x_0})\). Setting \(y=0\) gives \(S_x=(2x_0,0)\). 4. Their midpoint is \(\left(\frac{2x_0}{2},\frac{8/x_0}{2}\right)=\left(x_0,\frac4{x_0}\right)=P\).

a) \(f'(x)=\frac{0\cdot x-4\cdot1}{x^2}=-\frac4{x^2}\) b) \(S_x=(2x_0,0)\), \(S_y=(0,\frac8{x_0})\), and their midpoint is \((x_0,\frac4{x_0})=P\).

- Use the quotient rule directly in part a. - In part b, simplify the logarithm with a power property before applying the quotient rule. - Keep the domain restriction in mind when using logarithm identities.

1. For part a, apply the quotient rule: \(g'(x)=\frac{\frac{1}{x}(x+1)-\ln x}{(x+1)^2}\). 2. Combining the numerator over \(x\) gives \(g'(x)=\frac{x+1-x\ln x}{x(x+1)^2}\). 3. For part b, since \(x>0\), \(\ln(x^2)=2\ln x\). Thus \(h(x)=\frac{2\ln x}{x}\). 4. Applying the quotient rule gives \(h'(x)=\frac{2-2\ln x}{x^2}\).

a) \(g'(x)=\frac{x+1-x\ln x}{x(x+1)^2}\) b) \(h'(x)=\frac{2-2\ln x}{x^2}\)

- Apply the quotient rule and factor the numerator. - Check that each critical point belongs to the domain. - Use a sign chart to classify the extrema. - Substitute each point into the line equation.

1. By the quotient rule, \(f_k'(x)=\frac{2x(x-k)-x^2}{(x-k)^2}=\frac{x(x-2k)}{(x-k)^2}\). 2. The critical points are \(x=0\) and \(x=2k\), both of which are in the domain because \(k\neq0\). Their coordinates are \((0, 0)\) and \((2k, 4k)\). 3. The derivative changes sign at each critical point, so both are local extrema. Their classifications depend on the sign of \(k\): for \(k>0\), \((0, 0)\) is a local maximum and \((2k, 4k)\) is a local minimum; for \(k<0\), the classifications are reversed. 4. Both points satisfy \(y=2x\): \(0=2(0)\) and \(4k=2(2k)\).

a) \((0, 0)\) and \((2k, 4k)\). For \(k>0\), they are a local maximum and local minimum, respectively; for \(k<0\), the classifications are reversed. b) \(y=2x\)

- Apply the quotient rule and factor the derivative. - Check that the critical points are in the domain. - Evaluate the function at each critical point. - Substitute each point into \(y=x^2\).

1. By the quotient rule, \(f_t'(x)=\frac{2tx(x-t)-tx^2}{(x-t)^2}=\frac{tx(x-2t)}{(x-t)^2}\). 2. The critical points are \(x=0\) and \(x=2t\). Both are in the domain because \(t\neq0\). 3. Their coordinates are \((0, 0)\) and \((2t, 4t^2)\). A sign chart for the derivative confirms that both are local extrema. 4. The first point satisfies \(0=0^2\), and the second satisfies \(4t^2=(2t)^2\). Therefore, every local extremum lies on \(y=x^2\).

The local extrema are \((0, 0)\) and \((2t, 4t^2)\), and both satisfy \(y=x^2\).

- Use the given point to obtain one equation in the parameters. - A horizontal tangent gives a condition on the first derivative. - Apply the quotient rule before substituting the tangent input.

1. The point condition gives \(f(4)=8\): \(\frac{16a+4b}{2}=8\), so \(4a+b=4\). 2. By the quotient rule, \(f'(x)=\frac{(2ax+b)(x-2)-(ax^2+bx)}{(x-2)^2}=\frac{ax^2-4ax-2b}{(x-2)^2}\). 3. A horizontal tangent at \(x=0\) requires \(f'(0)=0\), so \(b=0\). 4. Substituting \(b=0\) into \(4a+b=4\) gives \(a=1\).

\(a=1\) and \(b=0\)

- Write the Quotient Rule setup before simplifying the derivative. - At a tangency point, the function values agree and the slopes agree. - Eliminate \(k\) to obtain an equation involving only the tangency input.

1. The quotient rule gives \(f'(x)=\frac{e^x x-e^x}{x^2}=\frac{e^x(x-1)}{x^2}\). 2. Let the tangency input be \(x_0\). Tangency requires \(f(x_0)=kx_0\) and \(f'(x_0)=k\). 3. Substitute \(k=f'(x_0)\): \(\frac{e^{x_0}}{x_0}=\frac{e^{x_0}(x_0-1)}{x_0^2}x_0\). 4. Since \(x_0\ne0\) and \(e^{x_0}>0\), this reduces to \(1=x_0-1\), so \(x_0=2\). 5. Therefore, \(k=f'(2)=\frac{e^2}{4}\), and the point of tangency is \((2,\frac{e^2}{2})\).

a) \(f'(x)=\frac{e^x x-e^x}{x^2}=\frac{e^x(x-1)}{x^2}\) b) \(k=\frac{e^2}{4}\); point of tangency \(\left(2,\frac{e^2}{2}\right)\)

- Differentiate the numerator and denominator separately before applying the quotient rule. - Keep the subtraction in the quotient-rule numerator grouped. - Substitute the requested input after simplifying the derivative.

1. Let \(u(x)=x^2-3x\) and \(v(x)=x+2\). Then \(u'(x)=2x-3\) and \(v'(x)=1\). 2. Apply the quotient rule: \(g'(x)=\frac{(2x-3)(x+2)-(x^2-3x)}{(x+2)^2}=\frac{x^2+4x-6}{(x+2)^2}\). 3. Evaluate at \(x=1\): \(g'(1)=-\frac{1}{9}\).

\(g'(1)=-\frac{1}{9}\)

- Find the point's y-coordinate before writing the tangent line. - Use the quotient rule to determine the tangent slope. - Convert the final positive slope to an inclination angle with inverse tangent.

1. Evaluate the function: \(k(2)=\frac{4}{3}\), so \(P=\left(2,\frac{4}{3}\right)\). 2. By the quotient rule, \(k'(x)=\frac{4x(x^2+2)-4x^3}{(x^2+2)^2}=\frac{8x}{(x^2+2)^2}\). 3. The tangent slope is \(m=k'(2)=\frac{4}{9}\). 4. Point-slope form gives \(y-\frac{4}{3}=\frac{4}{9}(x-2)\), so \(y=\frac{4}{9}x+\frac{4}{9}\). 5. The inclination angle satisfies \(\tan(\alpha)=\frac{4}{9}\), so \(\alpha\approx23.96^\circ\).

The tangent line is \(y=\frac{4}{9}x+\frac{4}{9}\), and its angle with the positive x-axis is approximately \(23.96^\circ\).

- Use the two displayed graphs to form a conjecture about the x-coordinate of the extremum, then prove it from the derivative. - Factor the derivative so its sign can be determined without a second derivative. - Compare the sign of \(1-\ln x\) on the two sides of \(e\), then account for the sign of \(a\).

1. The two displayed positive-\(a\) graphs suggest that their extrema occur at the same x-coordinate. 2. a) Substituting \(P(e^2,2)\) gives \(\frac{2a}{e^2}=2\), so \(a=e^2\). 3. b) The quotient rule gives \(f_a'(x)=a\frac{1-\ln x}{x^2}\). Since \(a\ne0\) and \(x^2>0\), the only critical point occurs when \(\ln x=1\), so \(x=e\). 4. The factor \(1-\ln x\) is positive for \(0<x<e\) and negative for \(x>e\). Therefore, if \(a>0\), \(f_a'\) changes from positive to negative and \(x=e\) is a local maximum. If \(a<0\), the signs reverse and \(x=e\) is a local minimum. 5. Every extremum has x-coordinate \(e\), so all extrema lie on the vertical line \(x=e\). 6. c) At the zero \(x=1\), \(f_a'(1)=a\).

a) \(a=e^2\) b) The only critical point is at \(x=e\); it is a local maximum for \(a>0\) and a local minimum for \(a<0\). All extrema lie on \(x=e\). c) \(m=a\)

- Use the point condition to obtain one equation in the parameters. - Use the quotient rule and the given tangent slope for a second equation. - Solve the resulting linear system only after simplifying the derivative condition.

1. Since \(P\) lies on the graph, \(4=\frac{4a+b}{3}\), so \(4a+b=12\). 2. By the quotient rule, \(f'(x)=\frac{2ax(x^2-1)-(ax^2+b)(2x)}{(x^2-1)^2}=\frac{-2x(a+b)}{(x^2-1)^2}\). 3. Use the slope condition: \(-\frac{8}{3}=f'(2)=-\frac{4(a+b)}{9}\), so \(a+b=6\). 4. Solving \(4a+b=12\) and \(a+b=6\) gives \(a=2\) and \(b=4\).

\(a=2\), \(b=4\)

- Use the vertical asymptote to determine the denominator parameter. - Use the x-intercept to relate the numerator coefficients. - Apply the quotient rule and then impose the given slope at the intercept.

1. The vertical asymptote at \(x=-2\) requires \(-2+c=0\), so \(c=2\). 2. Since \(x=2\) is an x-intercept, \(4a+2b=0\), so \(b=-2a\). Thus \(f(x)=\frac{ax^2-2ax}{x+2}\). 3. By the quotient rule, \(f'(x)=\frac{(2ax-2a)(x+2)-(ax^2-2ax)}{(x+2)^2}=\frac{ax^2+4ax-4a}{(x+2)^2}\). 4. At \(x=2\), \(f'(2)=\frac{a}{2}\). Since \(f'(2)=0.5\), \(a=1\). 5. Therefore, \(b=-2\) and \(c=2\), giving \(f(x)=\frac{x^2-2x}{x+2}\).

\(a=1\), \(b=-2\), and \(c=2\); therefore, \(f(x)=\frac{x^2-2x}{x+2}\).

- Divide each polygon into \(n\) congruent triangles from the center. - Use the radius and central angle to find each triangular area. - Apply sine and tangent in the relevant right triangles. - For monotonicity, use the quotient rule to differentiate \(\frac{\sin x}{x}\) and \(\frac{\tan x}{x}\), then determine the signs.

1. Divide the inscribed polygon into \(n\) isosceles triangles with side lengths \(R\) and central angle \(\frac{2\pi}{n}\). Each triangle has area \(\frac12R^2\sin\left(\frac{2\pi}{n}\right)\). Therefore, \(A_{i,n}=\frac n2R^2\sin\left(\frac{2\pi}{n}\right)\). 2. For the circumscribed polygon, \(R\) is the apothem. Each half-side has length \(R\tan\left(\frac{\pi}{n}\right)\), so each triangular section has area \(R^2\tan\left(\frac{\pi}{n}\right)\). Thus, \(A_{c,n}=nR^2\tan\left(\frac{\pi}{n}\right)\). 3. Rewrite the inscribed area as \(A_{i,n}=\pi R^2\frac{\sin(2\pi/n)}{2\pi/n}\). Since \(2\pi/n\to0\), \(A_{i,n}\to\pi R^2\). 4. Rewrite the circumscribed area as \(A_{c,n}=\pi R^2\frac{\tan(\pi/n)}{\pi/n}=\pi R^2\frac{\sin(\pi/n)}{(\pi/n)\cos(\pi/n)}\). Since \(\sin x/x\to1\) and \(\cos x\to1\), \(A_{c,n}\to\pi R^2\). 5. For \(f(x)=\frac{\sin x}{x}\), \(f'(x)=\frac{x\cos x-\sin x}{x^2}<0\) on \(0<x\le\frac{2\pi}{3}\). Because \(2\pi/n\) decreases as \(n\) increases, \(A_{i,n}\) is strictly increasing. 6. For \(g(x)=\frac{\tan x}{x}\), \(g'(x)=\frac{x\sec^2x-\tan x}{x^2}>0\) on \(0<x\le\frac{\pi}{3}\). Because \(\pi/n\) decreases as \(n\) increases, \(A_{c,n}\) is strictly decreasing.

a) \(A_{i,n}=\frac n2R^2\sin\left(\frac{2\pi}{n}\right)\) and \(A_{c,n}=nR^2\tan\left(\frac{\pi}{n}\right)\). b) Both sequences converge to \(\pi R^2\). c) For \(n\ge3\), \(A_{i,n}\) is strictly increasing and \(A_{c,n}\) is strictly decreasing.

- Track the power of the denominator factor after all cancellations, rather than assigning a special “order” name to the asymptote. - Factor powers of \(x-2\) before simplifying the derivative. - In the general case, evaluate the simplified numerator at \(x=a\) to decide whether cancellation is possible.

1. The denominator is zero at \(x=2\), while the numerator equals \(5\), so \(x=2\) is a vertical asymptote and the uncanceled denominator factor is \((x-2)^3\). 2. By the quotient rule, \(f'(x)=\frac{2x(x-2)^3-3(x^2+1)(x-2)^2}{(x-2)^6}\). Factoring and simplifying gives \(f'(x)=\frac{-x^2-4x-3}{(x-2)^4}\). 3. The numerator of \(f'\) equals \(-15\) at \(x=2\), so no further cancellation occurs. The denominator factor therefore has multiplicity \(4\). 4. For the general function, the quotient rule gives \(g'(x)=\frac{u'(x)(x-a)^n-nu(x)(x-a)^{n-1}}{(x-a)^{2n}}\). 5. Factoring and simplifying gives \(g'(x)=\frac{u'(x)(x-a)-nu(x)}{(x-a)^{n+1}}\). 6. At \(x=a\), the numerator equals \(-nu(a)\ne0\), so the denominator factor \((x-a)^{n+1}\) does not cancel.

1. Vertical asymptote: \(x=2\); denominator-factor multiplicity \(3\) 2. \(f'(x)=\frac{-x^2-4x-3}{(x-2)^4}\), with denominator-factor multiplicity \(4\) at \(x=2\) 3. \(g'(x)=\frac{u'(x)(x-a)-nu(x)}{(x-a)^{n+1}}\), and the factor \((x-a)^{n+1}\) is uncanceled because \(u(a)\ne0\)

- 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 \(a\).

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

- Use the quotient rule. - The denominator of the derivative is a square. Determine when the numerator is negative. - Compare the leading terms to evaluate the limit at infinity. - For a linear-over-linear rational function, compare the degrees of the numerator and denominator.

1. Apply the quotient rule: \(g_a'(x) = \frac{a(x + 2) - (ax + 4)}{(x + 2)^2} = \frac{2a - 4}{(x + 2)^2}\). 2. For \(x \ne -2\), the denominator \((x + 2)^2\) is positive. Therefore, \(g_a'(x) < 0\) exactly when \(2a - 4 < 0\), or \(a < 2\). When \(a = 2\), the function is constant on its domain. 3. Divide the numerator and denominator by \(x\): \(\lim_{x \to \infty} \frac{a + 4/x}{1 + 2/x} = a\). 4. No. For any linear-over-linear rational function with \(c \ne 0\), the limits as \(x \to \pm\infty\) are both the finite value \(\frac{a}{c}\). Its end behavior approaches the horizontal asymptote \(y = \frac{a}{c}\), not positive or negative infinity.

1) \(g_a'(x) = \frac{2a - 4}{(x + 2)^2}\) 2) \(a < 2\) 3) \(\lim_{x \to \infty} g_a(x) = a\) 4) No. A linear-over-linear rational function approaches the finite horizontal asymptote \(y = \frac{a}{c}\).

- 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.

- Recall how polynomial degrees behave under multiplication. - Determine how differentiation changes the degree of a nonconstant polynomial. - Compare the leading terms of \(u'v\) and \(uv'\). - Use the assumption \(n\ne m\) to decide whether those leading terms cancel.

1. Since \(u\) has degree \(n\), \(u'\) has degree \(n-1\). Therefore, \(\deg(u'v)=(n-1)+m=n+m-1\). Similarly, \(v'\) has degree \(m-1\), so \(\deg(uv')=n+(m-1)=n+m-1\). 2. Let \(a\) and \(b\) be the leading coefficients of \(u\) and \(v\). The leading terms of \(u'v\) and \(uv'\) are \(nabx^{n+m-1}\) and \(mabx^{n+m-1}\). Thus, the leading coefficient of \(u'v-uv'\) is \((n-m)ab\). Because \(n\ne m\), this coefficient is nonzero. Therefore, \(k=n+m-1\). 3. Since \(m\ge 1\), \(k-n=(n+m-1)-n=m-1\ge 0\). Hence, \(k\ge n\). This degree statement concerns the quotient-rule numerator before any common factor is canceled with the denominator.

1. Both terms have degree \(n+m-1\). 2. \(k=n+m-1\), because the leading coefficient is \((n-m)ab\ne 0\). 3. \(k-n=m-1\ge 0\), so \(k\ge n\).

- The vertical asymptote identifies the denominator zero. - Use the x-intercept to relate \(a\) and \(b\). - Differentiate with the quotient rule and apply the given slope. - Solve the resulting equations one parameter at a time.

1. The vertical asymptote is \(x=1\), so \(d=1\). Thus, \(f(x)=\frac{ax^2+bx}{x-1}\). 2. Since \(f(3)=0\), the numerator is zero at \(x=3\): \(9a+3b=0\). Therefore, \(b=-3a\), and \(f(x)=\frac{a(x^2-3x)}{x-1}\). 3. Differentiate using the quotient rule: \(f'(x)=a\frac{(2x-3)(x-1)-(x^2-3x)}{(x-1)^2}=a\frac{x^2-2x+3}{(x-1)^2}\). 4. The slope condition gives \(f'(3)=\frac{3}{2}\). Since \(f'(3)=a\frac{9-6+3}{4}=\frac{3a}{2}\), it follows that \(a=1\). 5. Therefore, \(b=-3\), and \(f(x)=\frac{x^2-3x}{x-1}\).

a) \(d=1\) b) \(a=1\), \(b=-3\) c) \(f(x)=\frac{x^2-3x}{x-1}\)

- Apply the quotient rule carefully. - Set the numerator of the derivative equal to zero. - Compare both coordinates of the maximum as \(k\) changes. - Compare the degrees of the numerator and denominator.

1. The quotient rule gives \(f_k'(x)=\frac{4(x^2+k)-8x^2}{(x^2+k)^2}=\frac{4k-4x^2}{(x^2+k)^2}\). 2. The derivative is zero at \(x=\sqrt{k}\). It changes from positive to negative, so this is a local maximum. Its value is \(f_k(\sqrt{k})=\frac{2}{\sqrt{k}}\), giving the local maximum \(\left(\sqrt{k}, \frac{2}{\sqrt{k}}\right)\). 3. As \(k\) increases, the maximum moves right and downward. Thus p corresponds to \(k=1\), q to \(k=4\), and r to \(k=9\). 4. The denominator has higher degree than the numerator, so \(\lim_{x\to\infty}f_k(x)=0\), independent of \(k\).

a) \(f_k'(x)=\frac{4k-4x^2}{(x^2+k)^2}\) b) \(\left(\sqrt{k}, \frac{2}{\sqrt{k}}\right)\) c) p: \(k=1\); q: \(k=4\); r: \(k=9\) d) \(0\), independent of \(k\)