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AP Calculus AB : Derivatives

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Example Questions

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Example Question #5 : Derivatives Of Functions

Find f'(x) :

(3x^3)(x)

Possible Answers:

f'(x)=9x^3

f'(x)=12x^2

f'(x)=9x^3+3x^2

f'(x)=12x^3

f'(x)=9x^2

Correct answer:

f'(x)=12x^3

Explanation:

This is the product rule, which is: (derivative of the first)(second)+(derivative of the second)(first)

So:

(9x^2)(x)+(1)(3x^3)=12x^3

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Example Question #21 : Ap Calculus Ab

Find the derivative of the following:

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

Possible Answers:

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

f'(x)=x-1

f'(x)=0

$f'(x)=x-\frac{1}{x}$

f'(x)=1

Correct answer:

f'(x)=1

Explanation:

This is a combination of chain rule and quotient rule.

So:

$2x(\frac{1}{x})+\frac{0-1}{x^2}(x^2)=2-\frac{x^2}{x^2}$

Which when simplified you get:

f'(x)=2-1=1

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Example Question #1 : Computation Of The Derivative

Find the derivative of the following:

f(x)=sin(x)+tan(x)

Possible Answers:

f'(x)=-cos(x)-sec^2(x)

f'(x)=cos(x)-sec^2(x)

f'(x)=-cos(x)+sec^2(x)

f'(x)=cos(x)+sec^2(x)

Correct answer:

f'(x)=cos(x)+sec^2(x)

Explanation:

This problem is just addition of derivatives using trigonometric functions.

So:

f'(x)=cos(x)+sec^2(x)

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Example Question #8 : Derivatives Of Functions

Find the derivative:

$\frac{tan(x)}{x^2}$

Possible Answers:

$f'(x)=\frac{-xsec^2(x)-2tan(x)}{x^3}$

$f'(x)=\frac{xsec^2(x)+2tan(x)}{x^3}$

$f'(x)=\frac{xsec^2(x)-2tan(x)}{x^3}$

$f'(x)=\frac{-xsec^2(x)+2tan(x)}{x^3}$

$f'(x)=\frac{xsec^2(x)-2tan(x)}{x^4}$

Correct answer:

$f'(x)=\frac{xsec^2(x)-2tan(x)}{x^3}$

Explanation:

The is a quotient rule using a trigonometric function.

So:

$f'(x)=\frac{x^2(sec^2(x))-tan(x)(2x)}{x^4}$

You can pull out an "x" and cancel it to get:

$f'(x)=\frac{xsec^2(x)-2tan(x)}{x^3}$

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Example Question #9 : Derivatives Of Functions

Find the derivative:

$f(x)=x^{-10}$

Possible Answers:

$f'(x)=\frac{10}{x^9}$

$f'(x)=\frac{-10}{x^9}$

$f'(x)=\frac{-10}{x^{11}}$

$f'(x)=\frac{10}{x^{11}}$

Correct answer:

$f'(x)=\frac{-10}{x^9}$

Explanation:

This is the same concept as a normal derivative just with a negative in the exponent.

$f'(x)=-10(x^{-9})$

which becomes:

$f'(x)=\frac{-10}{x^9}$

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Example Question #10 : Derivatives Of Functions

Calculate f'(x) :

$\small f'(x)={(x^2)}^2$

Possible Answers:

$\small f'(x)=3x^2$

$\small f'(x)=4x^2$

$\small f'(x)=2(x^2)(2x)$

$\small f'(x)=(x^2)(2x)$

Correct answer:

$\small f'(x)=2(x^2)(2x)$

Explanation:

This is a power rule that can utilize u-substitution.

So $\small f'(x)=(u)^2*u'$

where $\small u=x^2$

So you get:

$\small f'(x)=2u*u'$

Plug "u" back in and you get:

$\small f'(x)=2(x^2)(2x)$

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Example Question #21 : Derivatives

Find the derivative of the following:

$\small f(x)=\small \frac{x^3+2x}{x^3}$

Possible Answers:

$\small \small \small f'(x)=\frac{2}{x^2}$

$\small \small \small \small \small f'(x)=\frac{-2}{x^3}$

$\small \small \small \small f'(x)=\frac{-2}{x^4}$

$\small \small \small f'(x)=\frac{-2}{x^2}$

$\small \small \small \small f'(x)=\frac{2}{x^3}$

Correct answer:

$\small \small \small \small \small f'(x)=\frac{-2}{x^3}$

Explanation:

The easiest way to approach this problem is to break it up into terms to get:

$\small \small f(x)= \frac{x^3}{x^3}+\frac{2x}{x^3}$

This simplifies to:

$\small \small f(x)= 1+\frac{2}{x^2}$

This then becomes a simple derivative in which you get:

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

Which after simplifying you get:

$\small \small \small \small \small f'(x)=\frac{-2}{x^3}$

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Example Question #22 : Ap Calculus Ab

Find the derivative:

$\small x^4+x-cos(x)$

Possible Answers:

$\small f'(x)=\small -4x^3+1+sin(x)$

$\small f'(x)=\small 4x^3+1+sin(x)$

$\small f'(x)=\small 4x^3+1+cos(x)$

$\small f'(x)=\small 4x^3-1+sin(x)$

$\small f'(x)=\small 4x^3+1-sin(x)$

Correct answer:

$\small f'(x)=\small 4x^3+1+sin(x)$

Explanation:

This is a derivative of sums with a trigonometric function thrown in there.

Upon simplifying you get:

$\small f'(x)=\small 4x^3+1+sin(x)$

Keeping in mind the derivative of cos(x) is -sin(x)

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Example Question #23 : Derivatives

Find f'(x) :

$\small f(x)=sin(x)tan(x)$

Possible Answers:

$\small f(x)=sin(x)(1+csc^2(x))$

$\small \small f(x)=sin(x)(1+sec^2(x))$

$\small f(x)=-sin(x)(1+sec^2(x))$

$\small f(x)=sin(x)(1-sec^2(x))$

$\small f(x)=sin(x)(1-csc^2(x))$

Correct answer:

$\small \small f(x)=sin(x)(1+sec^2(x))$

Explanation:

This is a product rule using trigonometric functions:

$\small cos(x)(tan(x))+sec^2(x)sin(x)$

This can be simplified further:

$\small {\color{Red} cos(x)}(\frac{sin(x)}{{\color{Red} cos(x)}})+sin(x)sec^2(x)$

What is in red cancels and you get:

$\small {sin(x)}+sin(x)sec^2(x)$

But you can take this one step further and pull out a sin(x) to get:

$\small f'(x)=\small f(x)=sin(x)(1+sec^2(x))$

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Example Question #24 : Derivatives

Differentiate. $f(x)=7-x^{110}$

Possible Answers:

$-110x^{109}$

$109x^{109}$

$-109x^{109}$

$-110x^{110}$

$110x^{109}$

Correct answer:

$-110x^{109}$

Explanation:

The derivative of any term that does not include an "x" is zero. Therefore, the derivative of 7 is 0. The derivative of the second term follows if $f(x)=x^{a}$ then $f'(x)=a\times x^{a-1}$ . The negative sign can be pulled out as a -1 coefficient before you differentiate.

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AP Calculus AB : Derivatives

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #5 : Derivatives Of Functions

Example Question #21 : Ap Calculus Ab

Example Question #1 : Computation Of The Derivative

Example Question #8 : Derivatives Of Functions

Example Question #9 : Derivatives Of Functions

Example Question #10 : Derivatives Of Functions

Example Question #21 : Derivatives

Example Question #22 : Ap Calculus Ab

Example Question #23 : Derivatives

Example Question #24 : Derivatives

All AP Calculus AB Resources

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