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AP Calculus AB : Derivatives of functions

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Example Questions

Example Question #51 : Ap Calculus Ab

Find the first derivative of the following function:

$f=\alpha \sin (\beta^2x)e^{\gamma x}$ ,

where $\alpha, \beta, \gamma$ are all constants.

Possible Answers:

$\alpha \cos(\beta^2x)e^{\gamma x}+\alpha \sin(\beta^2x)e^{\gamma x}$

$\alpha \beta^2\cos(\beta^2x)e^{\gamma x}+\alpha \sin(\beta^2x)e^{\gamma x}$

$-\alpha \beta^2\cos(\beta^2x)e^{\gamma x}+\alpha \gamma \sin(\beta^2x)e^{\gamma x}$

$\alpha \cos(\beta^2x)e^{\gamma x}+\alpha \gamma \sin(\beta^2x)e^{\gamma x}$

$\alpha \beta^2\cos(\beta^2x)e^{\gamma x}+\alpha \gamma \sin(\beta^2x)e^{\gamma x}$

Correct answer:

$\alpha \beta^2\cos(\beta^2x)e^{\gamma x}+\alpha \gamma \sin(\beta^2x)e^{\gamma x}$

Explanation:

The derivative of the function is equal to

$f'=\alpha \beta^2\cos(\beta^2x)e^{\gamma x}+\alpha \gamma \sin(\beta^2x)e^{\gamma x}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{du} }{\mathrm{d} x}$

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Example Question #52 : Ap Calculus Ab

Find the second derivative of the following function:

$f=\cos(2x)e^{2x}$

Possible Answers:

$8\sin(2x)e^{2x}$

$-8e^{2x}(-\sin(2x)+\cos(2x))$

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

$-8\sin(2x)e^{2x}$

Correct answer:

$-8\sin(2x)e^{2x}$

Explanation:

To find the second derivative of the function, we first must find the first derivative, which is equal to

$f'=2e^{2x}(\sin(2x)+\cos(2x))$

which was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

The second derivative is equal to

$f''=-4\cos(2x)e^{2x}-4\sin(2x)e^{2x}-4\sin(2x)e^{2x}+4\cos(2x)e^{2x}=-8 \sin(2x)e^{2x}$

and was found using the same rules as above, as well as

$\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$

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Example Question #41 : Derivatives Of Functions

Find the second derivative of the following function:

$f=x^3+\sec(x)$

Possible Answers:

$3x^2+\sec(x)\tan(x)$

$6x+\sec(x)\tan^2(x)+\sec^3(x)$

$6x+\sec(x)\tan^2(x)-\sec^3(x)$

$6x-\sec(x)\tan^2(x)+\sec^3(x)$

Correct answer:

$6x+\sec(x)\tan^2(x)+\sec^3(x)$

Explanation:

First, we find the first derivative:

$f'=3x^2+\sec(x)\tan(x)$

This was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)$

Next, find the second derivative:

$f''=6x+\sec(x)\tan^2(x)+\sec^3(x)$

The following additional rules were used:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$

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Example Question #42 : Derivatives Of Functions

Find the first derivative of the following function:

$f(x)=x^2e^x+xe^{2x}$

Possible Answers:

e^x(e^x+2xe^x)

e^x(2x+x^2+e^x+2xe^x)

e^x(2x+x^2)

xe^x(2+x+3e^x)

Correct answer:

e^x(2x+x^2+e^x+2xe^x)

Explanation:

The first derivative is equal to the following:

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

which simplifies to

e^x(2x+x^2+e^x+2xe^x)

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u =e^u \frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x}ax=a$

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Example Question #61 : Derivatives

Given the function $\small \small \small \small \small z(x)=x^5-x^6-100$ , find its derivative.

Possible Answers:

$\small z'(x)=5x^4-6x^5$

$\small \small z'(x)=5x^4-6x^5-100$

$\small \small z'(x)=-5x^4+6x^5$

$\small \small z'(x)=5x^4-6x^5+100$

Correct answer:

$\small z'(x)=5x^4-6x^5$

Explanation:

Given the function $\small \small \small \small \small z(x)=x^5-x^6-100$ , we can find its derivative using the power rule, which states that

$\small \small [x^n]'=nx^{n-1}$

So we have

$\small z'(x)=5x^4-6x^5$

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Example Question #62 : Derivatives

Given the function $\small \small \small \small \small \small z(x)=x^\pi-x^{100}$ , find its derivative.

Possible Answers:

$\small \small \small z'(x)=\pi x^{\pi}-100x^{99}$

$\small \small z'(x)=\pi x^{\pi-1}-100x^{99}$

$\small \small \small z'(x)=\pi x^{\pi+1}-100x^{99}$

$\small \small \small z'(x)=\pi x^{\pi-1}-x^{100}$

Correct answer:

$\small \small z'(x)=\pi x^{\pi-1}-100x^{99}$

Explanation:

Given the function $\small \small \small \small \small \small z(x)=x^\pi-x^{100}$ , we can find its derivative using the power rule, which states that

$\small \small [x^n]'=nx^{n-1}$

So we have

$\small \small z'(x)=\pi x^{\pi-1}-100x^{99}$

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Example Question #63 : Derivatives

Given the function $\small \small \small \small \small \small \small z(x)=ex^5+\pi x^2$ , find its derivative.

Possible Answers:

$\small \small \small \small z'(x)=5ex^4+2\pi$

$\small \small \small \small z'(x)=5ex^5+2\pi x$

$\small \small \small \small z'(x)=5e^{5x}+2\pi x$

$\small \small \small z'(x)=5ex^4+2\pi x$

Correct answer:

$\small \small \small z'(x)=5ex^4+2\pi x$

Explanation:

Given the function $\small \small \small \small \small \small \small z(x)=ex^5+\pi x^2$ , we can find its derivative using the power rule, which states that

$\small \small [x^n]'=nx^{n-1}$

So we have

$\small \small \small z'(x)=5ex^4+2\pi x$

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Example Question #64 : Derivatives

Find the first derivative of the function:

$f(x)=\cos(x^2)+\tan(x^3)+x^4$

Possible Answers:

$-\sin(x^2)+\sec^2(x^3)+4x^3$

$2x\sin(x^2)+3x^2\sec^2(x^3)+4x^3$

$-2x\sin(x^2)+3x^2\sec^2(x^3)+4x^3$

$-2x\sin(x^2)+3x^2\sec^2(x^3)+4x$

Correct answer:

$-2x\sin(x^2)+3x^2\sec^2(x^3)+4x^3$

Explanation:

The derivative of the function is equal to

$f'=-2x\sin(x^2)+3x^2\sec^2(x^3)+4x^3$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$

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Example Question #61 : Derivatives

Find the derivative.

$3x^3-2x^\frac{1}{2}$

Possible Answers:

6x^2-x

$9x^2-\frac{1}{x^\frac{1}{2}}$

$6x^2-\frac{1}{x^\frac{1}{2}}$

9x^2-x

Correct answer:

$9x^2-\frac{1}{x^\frac{1}{2}}$

Explanation:

Use the power rule to find the derivative.

$\frac{d}{dx}3x^3=9x^2$

$\frac{d}{dx}-2x^\frac{1}{2}=-\frac{1}{x^\frac{1}{2}}$

Thus, the derivative is $9x^2-\frac{1}{x^\frac{1}{2}}$ .

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Example Question #62 : Derivatives

Find the derivative.

x^4-3x+24

Possible Answers:

4x-3

4x^3-3

x^3-x

4x^3+18

Correct answer:

4x^3-3

Explanation:

Use the power rule to find the derivative.

$\frac{d}{dx}x^4=4x^3$

$\frac{d}{dx}-3x=-3$

$\frac{d}{dx}24=0$

Thus, the derivative is 4x^3-3 .

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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 #51 : Ap Calculus Ab

Example Question #52 : Ap Calculus Ab

Example Question #41 : Derivatives Of Functions

Example Question #42 : Derivatives Of Functions

Example Question #61 : Derivatives

Example Question #62 : Derivatives

Example Question #63 : Derivatives

Example Question #64 : Derivatives

Example Question #61 : Derivatives

Example Question #62 : Derivatives

All AP Calculus AB Resources

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