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AP Calculus AB : AP Calculus AB

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3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Derivative Rules For Sums, Products, And Quotients Of Functions

If $f(x)=\pi\sin(ex)$ , evaluate f'(0) .

Possible Answers:

$\pi$

$e\pi$

Correct answer:

$e\pi$

Explanation:

When evaluating the derivative, pay attention to the fact that $e,\pi$ are constants, (not variables) and are treated as such.

$f'(x)=\pi\cos(ex)\times (ex)'=e\pi\cos(ex)$ .

and hence

$f'(0)=e\pi\cos(e(0))=e\pi$ .

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Example Question #2 : Derivative Rules For Sums, Products, And Quotients Of Functions

If f(x)=1+x+x^2+x^3+x^4 , evaluate f'(2)

Possible Answers:

Correct answer:

Explanation:

To obtain an expression for f'(x) , we can take the derivative of f(x) using the sum rule.

f'(x)=0+1+2x+3x^2+4x^3 .

Substituting into this equation gives us

f'(2) =0+1+4+12+32

=49 .

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Example Question #6 : Derivative Rules For Sums, Products, And Quotients Of Functions

If $f(x)=\frac{x\tan^{-1}(x)}{\ln(x)}$ , find f'(x) .

Possible Answers:

$\frac{\ln(x)(\frac{1}{1+x^2}+\tan^{-1}x)-\tan^{-1}x}{\ln(x^2)}$

$\frac{\ln(x)(\frac{1}{1+x^2}+\tan^{-1}x)-\tan^{-1}x}{(\ln(x))^2}$

$\frac{\frac{1}{x}(\frac{x}{1+x^2}+\tan^{-1}x)-\tan^{-1}x}{\ln(x^2)}$

$\frac{\ln(x)(\frac{x}{1+x^2}+\tan^{-1}x)-\tan^{-1}x}{(\ln(x))^2}$

$\frac{\ln(x)(\frac{x+1}{1+x^2})-\tan^{-1}x}{\ln(x^2)}$

Correct answer:

$\frac{\ln(x)(\frac{x}{1+x^2}+\tan^{-1}x)-\tan^{-1}x}{(\ln(x))^2}$

Explanation:

To find f'(x) , we will need to use the quotient rule; $\frac{gf'-fg'}{g^2}$ .

$f(x)= \frac{x\tan^{-1}x}{\ln(x)}$ . Start

$f'(x)= \frac{(\ln(x))(x\tan^{-1}x)'-(x\tan^{-1}x)(\ln(x))'}{(\ln(x))^2}$ . Use the quotient rule.

$= \frac{(\ln(x))(x(\frac{1}{1+x^2})+1(\tan^{-1}x))-(x\tan^{-1}x)(\frac{1}{x})}{(\ln(x))^2}$ . Take the derivatives inside of the quotient rule. The derivative of $x\tan^{-1}x$ uses the product rule.

$=\frac{\ln(x)(\frac{x}{1+x^2}+\tan^{-1}x)-\tan^{-1}x}{(\ln(x))^2}$ . Simplify to match the correct answer.

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Example Question #1 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative of the following equation:

$h(x) = \sin(x)\cos(x)$

Possible Answers:

$\cos ^{2}(x)+\sin^{2}(x)$

$-\sin(x)\cos(x)$

$\cos ^{2}(x)$

$\cos^{2}(x)-\sin^{2}(x)$

None of the other answers

Correct answer:

$\cos^{2}(x)-\sin^{2}(x)$

Explanation:

Because this problem contains two functions multiplied together that can't be simplified any further, it calls for the product rule, which states that $\frac{d}{dx}(f(x)g(x)) = f{}'(x)g(x) + f(x)g{}'(x)$ .

By using this rule, we get the answer:

$\cos(x)\cos(x) + \sin(x)(-\sin (x))$

By simplifying, we conclude that the derivative is equal to

$\cos ^{2}(x)-\sin ^{2}(x)$ .

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Example Question #8 : Derivative Rules For Sums, Products, And Quotients Of Functions

$\begin{align*}&\text{Calculate the derivative of the function}\\&-\frac{(114cos(19x))}{x^{20}}\end{align*}$

Possible Answers:

$-\frac{ (43320cos(19x))}{x^{21}}-\frac{ (43320sin(19x))}{x^{20}}$

$\frac{120}{x^{21}}- 361sin(19x)$

$\frac{(2280cos(19x))}{x^{21}}+\frac{ (2166sin(19x))}{x^{20}}$

$-\frac{(43320sin(19x))}{x^{21}}$

Correct answer:

$\frac{(2280cos(19x))}{x^{21}}+\frac{ (2166sin(19x))}{x^{20}}$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }cos(19x)\\&\text{and the second being: }\frac{1}{x^{20}}\\&\text{With the product scaled by }-114\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[cos(ax)]=-asin(ax)dx\\&d[x^a]=ax^{a-1}dx\\&\text{From the above we can continue:}\\&u=cos(19x);du=-19sin(19x)\\&v=\frac{1}{x^{20}};dv=-\frac{20}{x^{21}}\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&\frac{(2280cos(19x))}{x^{21}}+\frac{ (2166sin(19x))}{x^{20}}\end{align*}$

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Example Question #9 : Derivative Rules For Sums, Products, And Quotients Of Functions

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=-\frac{(380tan(17x))}{2^{(20x)}}\end{align*}$

Possible Answers:

$\frac{(380ln(2))}{2^{(20x)}}+ 340tan(17x)^{2} + 340$

$\frac{(7600\cdot (17tan(17x)^{2} + 17))}{2^{(20x)}}+\frac{ (129200tan(17x)ln(2))}{2^{(20x)}}$

$\frac{(7600tan(17x)ln(2))}{2^{(20x)}}-\frac{ (380\cdot (17tan(17x)^{2} + 17))}{2^{(20x)}}$

$\frac{(20ln(2)\cdot (6460tan(17x)^{2} + 6460))}{2^{(20x)}}$

Correct answer:

$\frac{(7600tan(17x)ln(2))}{2^{(20x)}}-\frac{ (380\cdot (17tan(17x)^{2} + 17))}{2^{(20x)}}$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }tan(17x)\\&\text{and the second being: }\frac{1}{2^{(20x)}}\\&\text{With the product scaled by }-380\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[tan(ax)]=\frac{adx}{cos^2(ax)}=(atan^2(ax)+a)dx\\&d[b^{ax}]=ab^{ax}ln(b)sx\\&\text{From the above we can continue:}\\&u=tan(17x);du=17tan(17x)^{2} + 17\\&v=\frac{1}{2^{(20x)}};dv=-\frac{(20ln(2))}{2^{(20x)}}\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&\frac{(7600tan(17x)ln(2))}{2^{(20x)}}-\frac{ (380\cdot (17tan(17x)^{2} + 17))}{2^{(20x)}}\end{align*}$

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Example Question #381 : Derivatives

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

Possible Answers:

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

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

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

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

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 #391 : Derivatives

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

Possible Answers:

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

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

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

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

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 #212 : Computation Of The Derivative

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

Possible Answers:

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

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

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

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

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 #213 : Computation Of The Derivative

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

Possible Answers:

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

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

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

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

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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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #1 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #2 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #6 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #1 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #8 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #9 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #381 : Derivatives

Example Question #391 : Derivatives

Example Question #212 : Computation Of The Derivative

Example Question #213 : Computation Of The Derivative

All AP Calculus AB Resources

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