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AP Calculus AB : Derivatives of functions

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Example Questions

Example Question #1 : Derivatives Of Functions

The Second Fundamental Theorem of Calculus (FTOC)

Consider the function equation (1)

$F(x)=\int_a^xf(t)dt$ (1)

The Second FTOC states that if:

is continuous on an open interval .
is in .
and is the anti derivative of

then we must have,

F'(x)=f(x) (2)

Differentiate,

$f(x)=\sin(x^2)+\int_o^{x^2+1}\ln(t+1)dt$

Possible Answers:

$f'(x)=2x\left[\cos(x^2)+\frac{1}{x^2+1}\right]$

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

$f'(x)=2x\cos(x^2)+\ln(x^2+1)$

$f'(x)=2x\left[\cos(x^2)+\ln(x^2+2)\right]$

Correct answer:

$f'(x)=2x\left[\cos(x^2)+\ln(x^2+2)\right]$

Explanation:

Differentiate:

$f(x)=\sin(x^2)+\int_o^{x^2+1}\ln(t+1)dt$

$f'(x) = \frac{d}{dx}\sin(x^2)+\frac{d}{dx}\int_o^{x^2+1} \ln(t+1)dt$

Both terms must be differentiated using the chain rule. The second term will use a combination of the chain rule and the Second Fundamental Theorem of Calculus. To make the derivative of the second term easier to understand, define a new variable so that the limits of integration will have the form shown in Equation (1) in the pre-question text.

Let,

u= x^2+1

Therefore,

$\frac{du}{dx}=2x$

Now we can write the derivative using the chain rule as:

$f'(x) = 2x\cos(x^2)+\frac{du}{dx}\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)$

$f'(x) = 2x\cos(x^2)+2x\underset{Use_, FTOC}{ \underbrace{\frac{d}{du}\left(\int_o^{u} \ln(t+1)dt\right)}}$

Let's calculate the derivative with respect to in the second term using the Second FTOC and then convert back to .

$\frac{d}{du}\int_o^{u} \ln(t+1)dt = \ln (u+1) = \ln\left \{ (x^2+1)+1 \right \} = \ln(x^2+2)$

Therefore we have,

$f'(x)=2x\cos(x^2)+2x\ln(x^2+2)$

$f'(x)=2x\left[\cos(x^2)+\ln(x^2+2)\right]$

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Example Question #13 : Derivatives

Find the derivative.

12x^3

Possible Answers:

36x^3

24x^2

36x^2

Correct answer:

36x^2

Explanation:

Use the power rule to find the derivative.

$\frac{d}{dx}12x^3=36x^2$

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Example Question #14 : Derivatives

Find the derivative of the following:

f(x)=3x+2

Possible Answers:

f'(x)=5

None of the above

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

f'(x)=3

Correct answer:

f'(x)=3

Explanation:

To take the derivative you need to bring the power down to the front of the equation, multiplying it by the coefficient and then drop the power.

So:

f(x)=3x+2 becomes f'(x)=3

because the degree of "x" is just one, and once you multiply 3 by 1 you get 3 and drop the power of "x" to 0. The second term is just a constant and the derivative of any term is just 0.

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Example Question #2 : Derivatives Of Functions

Find f'(x) :

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

Possible Answers:

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

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

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

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

f'(x)=0

Correct answer:

f'(x)=0

Explanation:

To do this problem you need to use the quotient rule. So you do

(low)(derivative of the high)-(high)(derivative of the low) all divided by the bottom term squared.

So:

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

Which, when simplified is:

$\frac{2x-2x}{x^2}=0$

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Example Question #21 : Derivatives

Find f'(x) :

(3x^3)(x)

Possible Answers:

f'(x)=9x^3

f'(x)=12x^2

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

f'(x)=9x^2

f'(x)=12x^3

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 #22 : Derivatives

Find the derivative of the following:

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

Possible Answers:

f'(x)=0

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

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

f'(x)=x-1

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 #23 : Derivatives

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 #24 : Ap Calculus Ab

Find the derivative:

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

Possible Answers:

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

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

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

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

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

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 #24 : Derivatives

Find the derivative:

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

Possible Answers:

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

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

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

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

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 #26 : Ap Calculus Ab

Calculate f'(x) :

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

Possible Answers:

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

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

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

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

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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AP Calculus AB : Derivatives of functions

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #1 : Derivatives Of Functions

Example Question #13 : Derivatives

Example Question #14 : Derivatives

Example Question #2 : Derivatives Of Functions

Example Question #21 : Derivatives

Example Question #22 : Derivatives

Example Question #23 : Derivatives

Example Question #24 : Ap Calculus Ab

Example Question #24 : Derivatives

Example Question #26 : Ap Calculus Ab

All AP Calculus AB Resources

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