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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 #731 : Ap Calculus Ab

Find the derivative of the function

$f=5x^2\tan{x}$

Possible Answers:

$f'=10x\tan{x}-5x^2\sec^2{x}$

$f'=10\tan{x}+5x^2\sec^2{x}$

$f'=10x\tan{x}+5x^2\sec^2{x}$

$f'=10x\tan{x}+5x^2\sec{x}$

Correct answer:

$f'=10x\tan{x}+5x^2\sec^2{x}$

Explanation:

To find the derivative of the function, you must apply the product rule. The product rule is as follows

$\frac{\mathrm{d} }{\mathrm{d} x}(f(x)g(x))=f'(x)g(x)+f(x)g'(x)$

In this case, we have f(x)=5x^2 and $g(x)=\tan{x}$

Using the product rule from above, we have

$f'=10x\tan{x}+5x^2\sec^2{x}$

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Example Question #42 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative.

$\frac{2x+1}{x^2}$

Possible Answers:

x-1

2x+1

$\frac{1}{x}$

Correct answer:

x-1

Explanation:

Use the quotient rule to find the derivative.

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

$=\frac{2x^2-4x^2-2x}{2x}$

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

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

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Example Question #245 : Computation Of The Derivative

Find the derivative.

$x^2\sin (x)$

Possible Answers:

$x^2\cos (x)+2x\cos (x)$

$x\cos (x)+2x\cos (x)$

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

$x\cos (x)+2x\sin (x)$

Correct answer:

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

Explanation:

Use the product rule to find the derivative.

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

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Example Question #733 : Ap Calculus Ab

Find the derivative of the function

f(x)=4x^3+10x

Possible Answers:

f'(x)=12x^2+10x

f'(x)=12x^2+10

f'(x)=12x^3+10

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

Correct answer:

f'(x)=12x^2+10

Explanation:

To find the derivative of a sum, we take the derivative of each part of the sum independently. Additionally, we apply the rules of differentiation that tell us

$\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

Differentiating the function from the problem statement, we get

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}(4x^3)+\frac{\mathrm{d} }{\mathrm{d} x}(10x)$

f'(x)=12x^2+10

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Example Question #734 : Ap Calculus Ab

Find the derivative of the function using the product and quotient rules:

$f(x)=\frac{5x\tan(x)}{\cos(x)}$

Possible Answers:

$f'(x)=(\cos(x)\left (5\tan(x)+5x\sec^2(x))+5x\tan(x)\sin(x))/\sin^2(x)$

$f'(x)=\cos(x)\left (5\tan(x)+5x\sec^2(x))+5x\tan(x)\sin(x)$

$f'(x)=(\cos(x)\left (5\tan(x)+5x\sec^2(x))+5x\tan(x)\sin(x))/\cos^2(x)$

$f'(x)=(\cos(x)\left (10\tan(x)+5\sec^2(x))+5x\tan(x)\sin(x))/\cos^2(x)$

Correct answer:

$f'(x)=(\cos(x)\left (5\tan(x)+5x\sec^2(x))+5x\tan(x)\sin(x))/\cos^2(x)$

Explanation:

Using the product and quotient rules to solve the problem, we get

$f'(x)=\frac{\cos(x)*\frac{\mathrm{d} }{\mathrm{d} x}(5x\tan(x))-(5x\tan(x))*\frac{\mathrm{d} }{\mathrm{d} x}(\cos(x))}{\cos^2(x)}$

The product rule is used when taking $\frac{\mathrm{d} }{\mathrm{d} x}(5x\tan(x))=\frac{\mathrm{d} }{\mathrm{d} x}(5x)*\tan(x)+5x*\frac{\mathrm{d} }{\mathrm{d} x}\tan(x)$

The final answer is

$f'(x)=(\cos(x)\left (5\tan(x)+5x\sec^2(x))+5x\tan(x)\sin(x))/\cos^2(x)$

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Example Question #735 : Ap Calculus Ab

Find the derivative of the function using the product rule:

f(x)=(x^3+3)(2x-1)

Possible Answers:

f'(x)=8x^3-3x^2+6

f'(x)=8x^2+3x^2+6

f'(x)=11x^3+4

f'(x)=8x+3x^2+5

Correct answer:

f'(x)=8x^3-3x^2+6

Explanation:

Using the product rule, we ge f'(x)=8x^3-3x^2+6 t

$f'(x)=(2x-1)*\frac{\mathrm{d} }{\mathrm{d} x}(x^3+3)+(x^3+3)*\frac{\mathrm{d} }{\mathrm{d} x}(2x-1)$

Simplifying,

f'(x)=3x^2(2x-1)+2(x^3+3)

f'(x)=6x^3-3x^2+2x^3+6

f'(x)=8x^3-3x^2+6

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Example Question #51 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative of y=5x^3e^x

Possible Answers:

15x^2e^x+5x^3xe

15x^2e^x+5x^3

none of these answers

5x^2e^x(3+x)

15x^2e^x

Correct answer:

5x^2e^x(3+x)

Explanation:

So whenever you have two distinct functions that are multiplied by each other, you will be using the product rule. So when looking at a function, see if you can separate it into two. In this case, we can see there is the function:

5x^3 and the function e^x .

So lets call those f(x) and g(x)

Then the product rule is that the derivative of f(x)g(x ) is:

f'(x)g(x)+f(x)g'(x) .

Then calculate to find:

f'(x)=15x^2 and g'(x)=e^x to give an answer of:

f'(x)g(x)+f(x)g'(x)=15x^2e^x+5x^3e^x

which simplifies to:

5x^2e^x(3+x)

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Example Question #52 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative of the following function:

f=2x^3+5x^2+x-4

Possible Answers:

f'=6x^2+11

f'=6x^2+10x+1

f'=6x^3+10x^2+x

f'=6x^2+10x+x-4

Correct answer:

f'=6x^2+10x+1

Explanation:

To find the derivative of the sum, we take the derivative of each term independently, then add them all up. Further, we use the rule

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

$f'=\frac{\mathrm{d} }{\mathrm{d} x}(2x^3)+\frac{\mathrm{d} }{\mathrm{d} x}(5x^2)+\frac{\mathrm{d} }{\mathrm{d} x}(x)+\frac{\mathrm{d} }{\mathrm{d} x}(-4)$

f'=6x^2+10x+1

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Example Question #53 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative of the following function

$f=5x\cos(x)$

Possible Answers:

$f'=5\cos(x)-5x\sin(x)$

$f'=5\cos(x)$

$f'=5\cos(x)+5\sin(x)$

$f'=5\cos(x)+5x\sin(x)$

Correct answer:

$f'=5\cos(x)-5x\sin(x)$

Explanation:

To find the derivative of the function, we must use the product rule, which is

$\frac{\mathrm{d} }{\mathrm{d} x}(f(x)*g(x))=f'(x)g(x)+f(x)g'(x)$

Using the function from the problem statement, we get

$f'=\frac{\mathrm{d} }{\mathrm{d} x}(5x)*(\cos(x))+\frac{\mathrm{d} }{\mathrm{d} x}(\cos(x))*5x$

$f'=5\cos(x)-5x\sin(x)$

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Example Question #251 : Computation Of The Derivative

Find the derivative of the following function:

$f=\frac{4x^3+\sin(x)}{\tan(x)}$

Possible Answers:

$f'=\frac{\tan(x)(12x^2+\cos(x))-(4x^3+\sin(x))(\sec^2(x))}{\tan^2(x)}$

$f'=\frac{\tan(x)(12x^3+\cos(x))-(4x^3+\cos(x))(\sec^2(x))}{\tan^2(x)}$

$f'=\frac{\tan(x)(12x^2+\cos(x))+(4x^3+\sin(x))(\tan^2(x))}{\tan^2(x)}$

$f'=\frac{\tan(x)(12x^2+\cos(x))-(4x^3+\sin(x))(\sec^2(x))}{\sec^2(x)}$

Correct answer:

$f'=\frac{\tan(x)(12x^2+\cos(x))-(4x^3+\sin(x))(\sec^2(x))}{\tan^2(x)}$

Explanation:

To find the derivative of the function, we use the quotient rule, which is

$\frac{\mathrm{d} }{\mathrm{d} x}(\frac{f}{g})=\frac{gf'-fg'}{g^2}$ , where and are any expression

Using the function from the problem statement, we get

$f'=\frac{\tan(x)*\frac{\mathrm{d} }{\mathrm{d} x}(4x^3+\sin(x))-(4x^3+\sin(x))*\frac{\mathrm{d} }{\mathrm{d} x}(\tan(x))}{\tan^2(x)}$

Taking the derivatives, we get

$f'=\frac{\tan(x)(12x^2+\cos(x))-(4x^3+\sin(x))(\sec^2(x))}{\tan^2(x)}$

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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 #731 : Ap Calculus Ab

Example Question #42 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #245 : Computation Of The Derivative

Example Question #733 : Ap Calculus Ab

Example Question #734 : Ap Calculus Ab

Example Question #735 : Ap Calculus Ab

Example Question #51 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #52 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #53 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #251 : Computation Of The Derivative

All AP Calculus AB Resources

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