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AP Calculus AB : Computation of the Derivative

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Example Questions

Example Question #467 : Derivatives

Find the derivative of $15x^5-2x^4+11x^2+e^x-\ln x$ .

Possible Answers:

$15x^4-2x^3+11x+e^x-\ln x$

55x^4-6x^3+22x

$25x^4+2x^3+11x+e^x+\ln x$

$75x^4-8x^3+22x+e^x-\tfrac{1} {x}$

$75x^4+8x^3-22x-e^x+\tfrac{1} {x}$

Correct answer:

$75x^4-8x^3+22x+e^x-\tfrac{1} {x}$

Explanation:

Take the derivative of each term.

$\tfrac{\mathrm{d} }{\mathrm{d} x}(15x^5)=75x^4$

$\tfrac{\mathrm{d} }{\mathrm{d} x}(-2x^4)=-8x^3$

$\tfrac{\mathrm{d} }{\mathrm{d} x}(11x^2)=22x$

$\tfrac{\mathrm{d} }{\mathrm{d} x}(e^x)=e^x$

$\tfrac{\mathrm{d} }{\mathrm{d} x}(-\ln x)=-\tfrac{1}{x}$

Add them:

$75x^4-8x^3+22x+e^x-\tfrac{1} {x}$

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Example Question #468 : Derivatives

Find the derivative of $2\sin x\cos x$ .

Possible Answers:

$2\cos2x$

$-2\cos x\sin x$

$2\sin2x$

$-\sin^2x+cos^2x$

Correct answer:

$2\cos2x$

Explanation:

There are two ways to solve this problem.

First, you can use a trig identity to replace $2\sin x\cos x$ with $\sin 2x$ . Using the chain rule, $(\sin 2x)'=2\cos2x$ .

Alternatively, you could use the product rule.

$2(\sin x(-\sin x)+\cos x(\cos x))=2(\cos^2x-\sin^2x)$

Since $\cos^2x-\sin^2x=\cos2x$ , our final answer is still $2\cos2x$ .

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Example Question #469 : Derivatives

Find the derivative of $\frac{e^x}{\sin x}$ .

Possible Answers:

$\frac{e^x}{\cos x}$

$\frac{e^x-e^x\tan x}{\sin x}$

$\frac{e^x\sin x-e^x\cos x}{\sin x}$

$\frac{e^x\tan x-e^x}{\cos x}$

$\frac{-e^x\tan x}{\sin x}$

Correct answer:

$\frac{e^x-e^x\tan x}{\sin x}$

Explanation:

For this problem, we need to use the quotient rule.

$(\frac{f}{g})'=\frac{gf'-fg'}{g^2}$

$(\frac{e^x}{\sin x})'=\frac{\sin x(e^x)-e^x(\cos x)}{\sin^2x}$

Simplifying:

$e^x (\frac{\sin x-\cos x}{\sin^2x})=\frac{e^x-e^x\tan x}{\sin x}$

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Example Question #91 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative of y=(2x^4+5x)(cos(x))

Possible Answers:

None of the other answers

y'=2x^4cos(x)-2x^4sin(x)+5xcos(x)-5xsin(x)

y'=-2x^4sin(x)+8x^3cos(x)-5xsin(x)+5cos(x)

y'=8x^3cos(x)+5cos(x)

Correct answer:

y'=-2x^4sin(x)+8x^3cos(x)-5xsin(x)+5cos(x)

Explanation:

$\frac{d}{dx}[2x^4+5x]=8x^3+5$

$\frac{d}{dx}[cos(x)]=-sin(x)$

Product rule states: $\frac{d}{du}[f(u) \cdot g(u)]=f'(u)g(u)+g'(u)f(u)$

Therefore:

y'=(8x^3+5)(cos(x))-sin(x)(2x^4+5x)

=-2x^4sin(x)+8x^3cos(x)-5xsin(x)+5cos(x)

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Example Question #92 : Derivative Rules For Sums, Products, And Quotients Of Functions

Use the chain rule to differentiate the following function: $h(x)=sin(x^{2}e^{x})$

Possible Answers:

$h^{'}(x)=-cos(x^{2}e^{x})xe^{x}(2+x)$

$h^{'}(x)=-sin(x^{2}e^{x})xe^{x}(2+x)$

$h^{'}(x)=cos(x^{2}e^{x})xe^{x}(2+x)$

$h^{'}(x)=sin(x^{2}e^{x})xe^{x}(2+x)$

Correct answer:

$h^{'}(x)=cos(x^{2}e^{x})xe^{x}(2+x)$

Explanation:

$h(x)=sin(x^{2}e^{x})$

By the chain rule: $h^{'}(x)=cos(x^{2}e^{x})\frac{\mathrm{d} }{\mathrm{d} x}(x^{2}e^{x})$

Differentiate $(x^{2}e^{x})$ using the product rule: $\frac{\mathrm{d} }{\mathrm{d} x}(x^{2}e^{x})=\frac{\mathrm{d} }{\mathrm{d} x}(x^{2})e^{x}+x^{2}\frac{\mathrm{d} }{\mathrm{d} x}e^{x}=2xe^{x}+x^{2}e^{x}$

Substitute this derivative for $\frac{\mathrm{d} }{\mathrm{d} x}(x^{2}e^{x})$ in the first equation: $h^{'}(x)=cos(x^{2}e^{x})(2xe^{x}+x^{2}e^{x})$

Factor the equation: $h^{'}(x)=cos(x^{2}e^{x})xe^{x}(2+x)$

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Example Question #93 : Derivative Rules For Sums, Products, And Quotients Of Functions

f(p)=cos(2p-1)-(2p^3+4p)

Find f'(p) .

Possible Answers:

f'(p)=sin(2p-1)-sin(2p^3+4p)

f'(p)=cos(2p-1)+sin(2p-1)-6p^2+4

f'(p)=-2cos(2p-1)(p^3+3p^2+2p+2)

f'(p)=-2(sin(2p-1)+3p^2+2)

f'(p) is undefined.

Correct answer:

f'(p)=-2(sin(2p-1)+3p^2+2)

Explanation:

$\frac{d}{dx}[cos(2p-1)]=2 \cdot -sin(2p-1)=-2sin(2p-1)$

$\frac{d}{dx}[2p^3+4p]=6p^2+4$

Therefore:

f'(p)=-2sin(2p-1)-(6p^2+4)

=-2sin(2p-1)-6p^2-4

=-2(sin(2p-1)+3p^2+2)

f'(p)=-2(3p^2+sin(2p-1)+2)

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Example Question #31 : Finding Derivatives

Let f(x)=sin(x)-cos(x) .

Find the second derivative of f(x) .

Possible Answers:

f''(x)=sin(x)-cos(x)

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

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

f''(x)=sin(x)+cos(x)

f''(x)=cos(x)-sin(x)

Correct answer:

f''(x)=cos(x)-sin(x)

Explanation:

The second derivative is just the derivative of the first derivative. So first we find the first derivative of f(x) . Remember the derivative of $\sin x$ is $\cos x$ , and the derivative for $\cos x$ is $-\sin x$ .

f'(x)= cos(x)+sin(x)

Then to get the second derivative, we just derive this function again. So

f''(x)=-sin(x)+cos(x)=cos(x)-sin(x)

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Example Question #51 : Calculus I — Derivatives

Define $f (x) = 6x^{3} - 12x^{2} + 4x -8$ .

What is f ' ' (x) ?

Possible Answers:

f ' '(x) = 36x

f ' '(x) = 36x - 24

f ''(x) = 18x - 24

f ' '(x) = 18

f ' '(x) = 6x - 1 2

Correct answer:

f ' '(x) = 36x - 24

Explanation:

Take the derivative of , then take the derivative of .

$f (x) = 6x^{3} - 12x^{2} + 4x -8$

$f '(x) = 3 \cdot 6x^{3-1} - 2 \cdot 12x^{2-1} + 4 -0$

$f '(x) = 18x^{2} - 24x+ 4$

$f ' '(x) = 2 \cdot 18x^{2-1} - 24+ 0$

f ' '(x) = 36x - 24

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Example Question #53 : Calculus I — Derivatives

Define $g(x) = \cos \left (x^{2} \right )$ .

What is g''(x) ?

Possible Answers:

$g''(x) = - 2 \cos(x^2)-4x^2 \sin(x^2) \right ]$

$g''(x) = -4x^2 \sin(x^2)$

$g''(x) =4x^2 \cos(x^2)$

$g''(x) =-4x^2 \cos(x^2)$

$g''(x) = - 2 \sin(x^2)-4x^2 \cos(x^2) \right ]$

Correct answer:

$g''(x) = - 2 \sin(x^2)-4x^2 \cos(x^2) \right ]$

Explanation:

Take the derivative of , then take the derivative of .

$g(x) = \cos \left (x^{2} \right )$

$g'(x) = 2x \cdot [-\sin (x^2)]$

$g'(x) = - 2x \sin (x^2)$

$g''(x) = - 2\left [ 1 \cdot \sin(x^2)+ x \cdot 2x \cos(x^2) \right ]$

$g''(x) = - 2\left [ \sin(x^2)+ 2x^2 \cos(x^2) \right ]$

$g''(x) = - 2 \sin(x^2)-4x^2 \cos(x^2) \right ]$

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Example Question #33 : Finding Derivatives

Define $f(x) = \frac{1}{e^{4x}}$ .

What is f ''(x) ?

Possible Answers:

$f''(x) = \frac{1 }{e^{4x+2}}$

$f(x) = \frac{1}{16e^{4x}}$

$f''(x) =- \frac{16 }{e^{4x}}$

$f''(x) = \frac{1 }{e^{4x-2}}$

$f''(x) = \frac{16 }{e^{4x}}$

Correct answer:

$f''(x) = \frac{16 }{e^{4x}}$

Explanation:

Rewrite:

$f(x) = \frac{1}{e^{4x}}$

$f(x) = e^{-4x}$

Take the derivative of , then take the derivative of .

$f'(x) = -4 e^{-4x}$

$f''(x) = -4 \cdot \left (-4 \right )e^{-4x} = 16e^{-4x} = \frac{16 }{e^{4x}}$

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AP Calculus AB : Computation of the Derivative

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #467 : Derivatives

Example Question #468 : Derivatives

Example Question #469 : Derivatives

Example Question #91 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #92 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #93 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #31 : Finding Derivatives

Example Question #51 : Calculus I — Derivatives

Example Question #53 : Calculus I — Derivatives

Example Question #33 : Finding Derivatives

All AP Calculus AB Resources

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