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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 #31 : Derivatives Of Functions

Given y(z), find y'(z).

y(z)=5e^z-16z^3+sin(z)

Possible Answers:

y'(z)=5e^z+32z^2+cos(z)

y'(z)=-5e^z-48z^2-cos(z)

y'(z)=5e^z-48z^2+cos(z)

y'(z)=5e^z-48z^2-sin(z)

Correct answer:

y'(z)=5e^z-48z^2+cos(z)

Explanation:

Given y(z), find y'(z).

y(z)=5e^z-16z^3+sin(z)

Now, we have three terms to our function, y(z). To find y'(z), we need to recall three rules.

1) $\frac{d}{dx}e^x=e^x$

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

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

Using these three rules, we can solve our problem.

1) $\frac{d}{dz}5e^z=5e^z$

2) $\frac{d}{dz}-16z^3=(3)(-16)z^{3-1}=-48z^2$

3) $\frac{d}{dz}(sin(z))=cos(z)$

Put all our terms together to get:

y'(z)=5e^z-48z^2+cos(z)

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Example Question #32 : Derivatives Of Functions

Given y(z), find y''(z).

y(z)=5e^z-16z^3+sin(z)

Possible Answers:

y''(z)=5e^z-96+sin(z)

y''(z)=5e^z-96z+sin(z)

y''(z)=-5e^z+96z-sin(z)

y''(z)=5e^z-96z-sin(z)

Correct answer:

y''(z)=5e^z-96z-sin(z)

Explanation:

Given y(z), find y''(z).

y(z)=5e^z-16z^3+sin(z)

Now, we have three terms to our function, y(z). To find y''(z), we need to first find y'(z), then we can differentiate again to get y"(z).

To find y'(z)...

1) $\frac{d}{dx}e^x=e^x$

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

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

Using these three rules, we can complete the first step.

1) $\frac{d}{dz}5e^z=5e^z$

2) $\frac{d}{dz}-16z^3=(3)(-16)z^{3-1}=-48z^2$

3) $\frac{d}{dz}(sin(z))=cos(z)$

Put all our terms together to get:

y'(z)=5e^z-48z^2+cos(z)

Now, we need to replicate the process to get y"(z).

1) Our first term will remain the same.

$\frac{d}{dz}5e^z=5e^z$

2) Our second term will follow the same rule and reduce again.

$\frac{d}{dz}-48z^2=(2)(-48)z^{2-1}=-96z$

3) Now we need to recall another rule:

$\frac{d}{dz}(cos(z))=-sin(z)$

So now we can put it all back together to get:

y''(z)=5e^z-96z-sin(z)

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Example Question #31 : Computation Of The Derivative

Find the derivative of the following function:

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

Possible Answers:

e^x(2x^2+4x)+2x+3

4xe^x+2x+3

e^x(2x^2+4x)+5x

2x^2e^x+2x+3

Correct answer:

e^x(2x^2+4x)+2x+3

Explanation:

The derivative of the function is equal to

f'(x)=e^x(2x^2+4x)+2x+3

and was found using the following rules:

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

For a sum, the derivative is simply the sum of the derivatives of the individual parts.

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Example Question #31 : Derivatives Of Functions

Find the first derivative of the following function:

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

Possible Answers:

$e^x(x^2+x)+12x^2\sec^2(4x^3)$

$e^x(x^2+2x)+12x^2\sec(4x^3)$

$2xe^x+12x^2\sec^2(4x^3)$

$e^x(x^2+2x)+12x^2\sec^2(4x^3)$

e^x(x^2+2x)+sec^2(4x^3)

Correct answer:

$e^x(x^2+2x)+12x^2\sec^2(4x^3)$

Explanation:

The first derivative of the function is equal to

$f'(x)=e^x(x^2+2x)+12x^2\sec^2(4x^3)$

and was found using the following rules:

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

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Example Question #35 : Derivatives Of Functions

Find the first derivative of the following function:

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

Possible Answers:

$\frac{2x+\tan(x)(x^2+3)}{\cos(x)}$

$\frac{2x+\sin(x)(x^2+3)}{\cos(x)}$

$\frac{2x\cos(x)+\tan(x)(x^2+3)}{\cos(x)}$

$\frac{-2x-\tan(x)(x^2+3)}{\cos(x)}$

Correct answer:

$\frac{2x+\tan(x)(x^2+3)}{\cos(x)}$

Explanation:

The derivative of the function is equal to

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

and was found using the following rules:

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

Taking the derivative without simplification gets us

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

and algebra is used to simplify to the final expression above.

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Example Question #36 : Derivatives Of Functions

Calculate the derivative:

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

Possible Answers:

$\small f'(x)=0$

$\small f'(x)=1$

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

$\small \small \small f'(x)=1-tan^2(x)$

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

Correct answer:

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

Explanation:

This is a chain rule using trigonometric functions.

$\small \small f'(x)=cos(x)sec(x)+sec(x)tan(x)sin(x)$

Which upon simplifying is:

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

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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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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #31 : Derivatives Of Functions

Example Question #32 : Derivatives Of Functions

Example Question #31 : Computation Of The Derivative

Example Question #31 : Derivatives Of Functions

Example Question #35 : Derivatives Of Functions

Example Question #36 : Derivatives Of Functions

Example Question #51 : Ap Calculus Ab

Example Question #52 : Ap Calculus Ab

Example Question #41 : Derivatives Of Functions

Example Question #42 : Derivatives Of Functions

All AP Calculus AB Resources

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