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AP Calculus AB : Derivative rules for sums, products, and quotients of functions

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Example Questions

Example Question #381 : Derivatives

$\begin{align*}&\text{Calculate the derivative of the function}\\&56sin(2x)tan(16x)\end{align*}$

Possible Answers:

$112sin(2x)\cdot (16tan(16x)^{2} + 16) + 1792cos(2x)tan(16x)$

$- 8cos(2x) - 224tan(16x)^{2} - 224$

$56sin(2x)\cdot (16tan(16x)^{2} + 16) + 112cos(2x)tan(16x)$

$2cos(2x)\cdot (896tan(16x)^{2} + 896)$

Correct answer:

$56sin(2x)\cdot (16tan(16x)^{2} + 16) + 112cos(2x)tan(16x)$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }tan(16x)\\&\text{and the second being: }sin(2x)\\&\text{With the product scaled by }56\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[tan(ax)]=\frac{adx}{cos^2(ax)}=(atan^2(ax)+a)dx\\&d[sin(ax)]=acos(ax)dx\\&\text{From the above we can continue:}\\&u=tan(16x);du=16tan(16x)^{2} + 16\\&v=sin(2x);dv=2cos(2x)\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&56sin(2x)\cdot (16tan(16x)^{2} + 16) + 112cos(2x)tan(16x)\end{align*}$

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Example Question #691 : Ap Calculus Ab

$\begin{align*}&\text{Find the derivative of the function }\\&f(x)=\frac{(136ln(17x))}{x^{8}}\end{align*}$

Possible Answers:

$\frac{136}{x^{9}}-\frac{ (1088ln(17x))}{x^{9}}$

$\frac{136}{x^{9}}-\frac{ 8}{x}$

$-\frac{1088}{x^{10}}$

$-\frac{ (18496ln(17x))}{x^{9}}-\frac{ 1088}{x^{9}}$

Correct answer:

$\frac{136}{x^{9}}-\frac{ (1088ln(17x))}{x^{9}}$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }\frac{1}{x^{8}}\\&\text{and the second being: }ln(17x)\\&\text{With the product scaled by }136\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[x^a]=ax^{a-1}dx\\&d[ln(ax)]=\frac{dx}{x}\\&\text{From the above we can continue:}\\&u=\frac{1}{x^{8}};du=-\frac{8}{x^{9}}\\&v=ln(17x);dv=\frac{1}{x}\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&\frac{136}{x^{9}}-\frac{ (1088ln(17x))}{x^{9}}\end{align*}$

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Example Question #392 : Derivatives

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=-182ln(16x)sin(4x)\end{align*}$

Possible Answers:

$-\frac{ (182sin(4x))}{x}- 728cos(4x)ln(16x)$

$-\frac{(728cos(4x))}{x}$ $-\frac{ (728sin(4x))}{x}- 11648cos(4x)ln(16x)$

$-\frac{(728cos(4x))}{x}$

$56cos(4x) -\frac{ 13}{x}$

Correct answer:

$-\frac{ (182sin(4x))}{x}- 728cos(4x)ln(16x)$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }ln(16x)\\&\text{and the second being: }sin(4x)\\&\text{With the product scaled by }-182\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[ln(ax)]=\frac{dx}{x}\\&d[sin(ax)]=acos(ax)dx\\&\text{From the above we can continue:}\\&u=ln(16x);du=\frac{1}{x}\\&v=sin(4x);dv=4cos(4x)\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&-\frac{ (182sin(4x))}{x}- 728cos(4x)ln(16x)\end{align*}$

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Example Question #391 : Derivatives

$\begin{align*}&\text{Calculate the derivative of the function}\\&-216\cdot 4^{(16x)}sin(12x)\end{align*}$

Possible Answers:

$192\cdot 4^{(16x)}ln(4) - 216cos(12x)$

$-41472\cdot 4^{(16x)}cos(12x)ln(4)$

$- 41472\cdot 4^{(16x)}cos(12x) - 41472\cdot 4^{(16x)}sin(12x)ln(4)$

$- 2592\cdot 4^{(16x)}cos(12x) - 3456\cdot 4^{(16x)}sin(12x)ln(4)$

Correct answer:

$- 2592\cdot 4^{(16x)}cos(12x) - 3456\cdot 4^{(16x)}sin(12x)ln(4)$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }4^{(16x)}\\&\text{and the second being: }sin(12x)\\&\text{With the product scaled by }-216\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[b^{ax}]=ab^{ax}ln(b)sx\\&d[sin(ax)]=acos(ax)dx\\&\text{From the above we can continue:}\\&u=4^{(16x)};du=16\cdot 4^{(16x)}ln(4)\\&v=sin(12x);dv=12cos(12x)\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&- 2592\cdot 4^{(16x)}cos(12x) - 3456\cdot 4^{(16x)}sin(12x)ln(4)\end{align*}$

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Example Question #394 : Derivatives

$\begin{align*}&\text{Find the derivative of the function }\\&f(x)=-119tan(6x)e^{(17x)}\end{align*}$

Possible Answers:

$12138tan(6x)e^{(17x)} - 2023e^{(17x)}\cdot (6tan(6x)^{2} + 6)$

$-2023e^{(17x)}\cdot (6tan(6x)^{2} + 6)$

$289e^{(17x)} - 42tan(6x)^{2} - 42$

$- 119e^{(17x)}\cdot (6tan(6x)^{2} + 6) - 2023tan(6x)e^{(17x)}$

Correct answer:

$- 119e^{(17x)}\cdot (6tan(6x)^{2} + 6) - 2023tan(6x)e^{(17x)}$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }e^{(17x)}\\&\text{and the second being: }-tan(6x)\\&\text{With the product scaled by }119\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[e^{ax}]=ae^{ax}dx\\&d[tan(ax)]=\frac{adx}{cos^2(ax)}=(atan^2(ax)+a)dx\\&\text{From the above we can continue:}\\&u=e^{(17x)};du=17e^{(17x)}\\&v=-tan(6x);dv=- 6tan(6x)^{2} - 6\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&- 119e^{(17x)}\cdot (6tan(6x)^{2} + 6) - 2023tan(6x)e^{(17x)}\end{align*}$

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Example Question #395 : Derivatives

$\begin{align*}&\text{Calculate the derivative of the function}\\&-\frac{(272cos(12x))}{3^{(12x)}}\end{align*}$

Possible Answers:

$\frac{(39168cos(12x)ln(3))}{3^{(12x)}}-\frac{ (39168sin(12x))}{3^{(12x)}}$

$-\frac{(39168sin(12x)ln(3))}{3^{(12x)}}$

$\frac{(204ln(3))}{3^{(12x)}}- 192sin(12x)$

$\frac{(3264sin(12x))}{3^{(12x)}}+\frac{ (3264cos(12x)ln(3))}{3^{(12x)}}$

Correct answer:

$\frac{(3264sin(12x))}{3^{(12x)}}+\frac{ (3264cos(12x)ln(3))}{3^{(12x)}}$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }\frac{1}{3^{(12x)}}\\&\text{and the second being: }cos(12x)\\&\text{With the product scaled by }-272\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[b^{ax}]=ab^{ax}ln(b)sx\\&d[cos(ax)]=-asin(ax)dx\\&\text{From the above we can continue:}\\&u=\frac{1}{3^{(12x)}};du=-\frac{(12ln(3))}{3^{(12x)}}\\&v=cos(12x);dv=-12sin(12x)\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&\frac{(3264sin(12x))}{3^{(12x)}}+\frac{ (3264cos(12x)ln(3))}{3^{(12x)}}\end{align*}$

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Example Question #396 : Derivatives

$\begin{align*}&\text{Find the derivative of the function }\\&f(x)=-112x^{17}ln(-4x)\end{align*}$

Possible Answers:

$- 1904x^{16}ln(-4x) - 112x^{16}$

$7616x^{16}ln(-4x) - 1904x^{16}$

$-1904x^{15}$

$119x^{16} -\frac{ 16}{x}$

Correct answer:

$- 1904x^{16}ln(-4x) - 112x^{16}$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }ln(-4x)\\&\text{and the second being: }x^{17}\\&\text{With the product scaled by }-112\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[ln(ax)]=\frac{dx}{x}\\&d[x^a]=ax^{a-1}dx\\&\text{From the above we can continue:}\\&u=ln(-4x);du=\frac{1}{x}\\&v=x^{17};dv=17x^{16}\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&- 1904x^{16}ln(-4x) - 112x^{16}\end{align*}$

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Example Question #391 : Derivatives

$\begin{align*}&\text{Calculate the derivative of the function}\\&130cos(6x)e^{(-14x)}\end{align*}$

Possible Answers:

$10920cos(6x)e^{(-14x)} + 10920sin(6x)e^{(-14x)}$

$10920sin(6x)e^{(-14x)}$

$- 182e^{(-14x)} - 60sin(6x)$

$- 1820cos(6x)e^{(-14x)} - 780sin(6x)e^{(-14x)}$

Correct answer:

$- 1820cos(6x)e^{(-14x)} - 780sin(6x)e^{(-14x)}$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }e^{(-14x)}\\&\text{and the second being: }cos(6x)\\&\text{With the product scaled by }130\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[e^{ax}]=ae^{ax}dx\\&d[cos(ax)]=-asin(ax)dx\\&\text{From the above we can continue:}\\&u=e^{(-14x)};du=-14e^{(-14x)}\\&v=cos(6x);dv=-6sin(6x)\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&- 1820cos(6x)e^{(-14x)} - 780sin(6x)e^{(-14x)}\end{align*}$

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Example Question #398 : Derivatives

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=-\frac{(220cos(3x))}{x^{16}}\end{align*}$

Possible Answers:

$\frac{(10560cos(3x))}{x^{17}}-\frac{ (10560sin(3x))}{x^{16}}$

$-\frac{(10560sin(3x))}{x^{17}}$

$33sin(3x) -\frac{ 320}{x^{17}}$

$\frac{(3520cos(3x))}{x^{17}}+\frac{ (660sin(3x))}{x^{16}}$

Correct answer:

$\frac{(3520cos(3x))}{x^{17}}+\frac{ (660sin(3x))}{x^{16}}$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }cos(3x)\\&\text{and the second being: }\frac{1}{x^{16}}\\&\text{With the product scaled by }-220\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[cos(ax)]=-asin(ax)dx\\&d[x^a]=ax^{a-1}dx\\&\text{From the above we can continue:}\\&u=cos(3x);du=-3sin(3x)\\&v=\frac{1}{x^{16}};dv=-\frac{16}{x^{17}}\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&\frac{(3520cos(3x))}{x^{17}}+\frac{ (660sin(3x))}{x^{16}}\end{align*}$

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Example Question #399 : Derivatives

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=32x^{17}tan(4x)\end{align*}$

Possible Answers:

$16tan(4x)^{2} + 136x^{16} + 16$

$2176x^{16}tan(4x) + 544x^{17}\cdot (4tan(4x)^{2} + 4)$

$544x^{16}\cdot (4tan(4x)^{2} + 4)$

$544x^{16}tan(4x) + 32x^{17}\cdot (4tan(4x)^{2} + 4)$

Correct answer:

$544x^{16}tan(4x) + 32x^{17}\cdot (4tan(4x)^{2} + 4)$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }x^{17}\\&\text{and the second being: }tan(4x)\\&\text{With the product scaled by }32\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[x^a]=ax^{a-1}dx\\&d[tan(ax)]=\frac{adx}{cos^2(ax)}=(atan^2(ax)+a)dx\\&\text{From the above we can continue:}\\&u=x^{17};du=17x^{16}\\&v=tan(4x);dv=4tan(4x)^{2} + 4\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&544x^{16}tan(4x) + 32x^{17}\cdot (4tan(4x)^{2} + 4)\end{align*}$

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AP Calculus AB : Derivative rules for sums, products, and quotients of functions

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #381 : Derivatives

Example Question #691 : Ap Calculus Ab

Example Question #392 : Derivatives

Example Question #391 : Derivatives

Example Question #394 : Derivatives

Example Question #395 : Derivatives

Example Question #396 : Derivatives

Example Question #391 : Derivatives

Example Question #398 : Derivatives

Example Question #399 : Derivatives

All AP Calculus AB Resources

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