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Calculus 2 : Derivative Review

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Example Questions

Example Question #241 : Derivatives

Find the velocity given the displacement function

$f(t)=10 (t-2)^{1/3}$

Possible Answers:

$f'(t)=\frac{10}{ 3(t-2)^{1/3}}$

$f'(t)=\frac{20}{ 3(t-2)^{2/3}}$

$f'(t)=\frac{10}{ 3(t-2)^{2/3}}$

$f'(t)=\frac{20}{ 3(t-2)^{1/3}}$

Correct answer:

$f'(t)=\frac{10}{ 3(t-2)^{2/3}}$

Explanation:

The derivative of the displacement function is the velocity, so we need to find f'(t) . We can use the power rule

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

with $a=\frac{1}{3}$

to get

$f'(t)=10\frac{1}{3}(t-2)^{-2/3}=\frac{10}{3(t-2)^{2/3}}$

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Example Question #242 : Derivatives

Find the velocity given the displacement function

$f(t)=5(3t-1)^{3}$

Possible Answers:

f'(t)=5(3t-1)^2

f'(t)=15(3t-1)^2

f'(t)=45(3t-1)^4

f'(t)=45(3t-1)^2

Correct answer:

f'(t)=45(3t-1)^2

Explanation:

The derivative of the displacement function is the velocity, so we need to find f'(t) . We can use the power rule

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

with a=3 along with the chain rule to get

$f'(t)=5(3t-1)^{2}\cdot 3\cdot 3=45(3t-1)^2$

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Example Question #43 : First And Second Derivatives Of Functions

Find the velocity given the displacement function

$f(t)=(t^3-1)^{2}$

Possible Answers:

f'(t)=6t^2(t^3-1)^2

f'(t)=3t^2(t^3-1)

f'(t)=6t^3(t^3-1)

f'(t)=6t^2(t^3-1)

Correct answer:

f'(t)=6t^2(t^3-1)

Explanation:

The derivative of the displacement function is the velocity, so we need to find f'(t) . We can use the power rule

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

with a=2 along with the chain rule to get

$f'(t)=(t^3-1)^{2}=3t^2\cdot 2(t^3-1)=6t^2(t^3-1)$

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Example Question #41 : First And Second Derivatives Of Functions

Find the velocity given the displacement function

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

Possible Answers:

$f'(x)=4e^x(e^{2x}-1)^3$

f'(x)=e^x(e^x-1)^3

f'(x)=4(e^x-1)^3

f'(x)=4e^x(e^x-1)^3

Correct answer:

f'(x)=4e^x(e^x-1)^3

Explanation:

The derivative of the displacement function is the velocity, so we need to find f'(x) . We can use the power rule

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

with a=4 along with the chain rule to get

$f'(x)=4\cdot e^x(e^x-1)^3$

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Example Question #45 : First And Second Derivatives Of Functions

Given the displacement function $f(x)=e^{20x+x^5}$ , find the velocity function.

Possible Answers:

$f'(x)=e^{20x+x^5}$

$f'(x)=(20+5x^4)e^{20+5x^4}$

$f'(x)=(20+5x^4)e^{20x+x^5}$

$f'(x)=e^{20+5x^4}$

Correct answer:

$f'(x)=(20+5x^4)e^{20x+x^5}$

Explanation:

To find the velocity of $f(x)=e^{20x+x^5}$ , we need to find the derivative. This can be done with the chain rule:

(g(h(x)))'=g'(h(x))h'(x)

with g(x)=e^x and h(x)=20x+x^5 , so we get

g'(x)=e^x

and

h'(x)=20+5x^4

so we get

$f'(x)=(20+5x^4)e^{20x+x^5}$

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Example Question #43 : First And Second Derivatives Of Functions

Given the displacement function $f(x)=\ln(e^x+e^{2x})$ , find the velocity function.

Possible Answers:

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

f'(x)=3

f'(x)=1

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

Correct answer:

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

Explanation:

To find the velocity of $f(x)=\ln(e^x+e^{2x})$ , we need to find the derivative. This can be done with the chain rule:

(g(h(x)))'=g'(h(x))h'(x)

with $g(x)=\ln x$ and $h(x)=e^x+e^{2x}$ , so we get

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

and

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

so we get

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

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Example Question #241 : Derivative Review

Given the displacement function

f(x)=ex+x^2

find the velocity function.

Possible Answers:

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

f'(x)=e+x^2

f'(x)=ex+2x

f'(x)=e+2x

Correct answer:

f'(x)=e+2x

Explanation:

To find the velocity of f(x)=ex+x^2 , we need to find the derivative. This is simply:

f'(x)=e+2x

using the power rule:

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

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Example Question #45 : First And Second Derivatives Of Functions

Given the velocity function

$v(x)=e^{5x}-e^{-5x}$

find the acceleration function a(x) .

Possible Answers:

$a(x)=e^{5x}-e^{-5x}$

$a(x)=5e^{5x}-5e^{-5x}$

$a(x)=5e^{5x}+5e^{-5x}$

Correct answer:

$a(x)=5e^{5x}+5e^{-5x}$

Explanation:

The acceleration function a(x) can be derived from the velocity function v(x) by taking the derivative: a(x)=v'(x) . So we get

$a(x)=v'(x)=\frac{d}{dx}(e^{5x}-e^{-5x})=5e^{5x}+5e^{-5x}$

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Example Question #46 : First And Second Derivatives Of Functions

Given the velocity function

$v(x)=e^{2x}-4e^{-6x}$

find the acceleration function a(x) .

Possible Answers:

$a(x)=e^{2x}+e^{-6x}$

$a(x)=2e^{2x}+24e^{-6x}$

$a(x)=2e^{2x}-24e^{-6x}$

Correct answer:

$a(x)=2e^{2x}+24e^{-6x}$

Explanation:

The acceleration function a(x) can be derived from the velocity function v(x) by taking the derivative: a(x)=v'(x) . So we get

$a(x)=v'(x)=\frac{d}{dx}(e^{2x}-4e^{-6x})=2e^{2x}+24e^{-6x}$

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Example Question #47 : First And Second Derivatives Of Functions

Find the derivative of $y= e^{-8x}\cos(x)$

Possible Answers:

$-8\sin(x)$

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

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

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

None of the other answers

Correct answer:

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

Explanation:

We need to use the Product Rule to evaluate the derivative here.

The formula for the Product Rule is

(fg)'(x) = f(x)g'(x)+f'(x)g(x)

If $f(x) = e^{-8x}$ , $g(x)=\cos(x)$ , then $f'(x)=-8e^{-8x}$ , $g'(x)=-\sin(x)$

Plugging these into our formula, we get-

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

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

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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #241 : Derivatives

Example Question #242 : Derivatives

Example Question #43 : First And Second Derivatives Of Functions

Example Question #41 : First And Second Derivatives Of Functions

Example Question #45 : First And Second Derivatives Of Functions

Example Question #43 : First And Second Derivatives Of Functions

Example Question #241 : Derivative Review

Example Question #45 : First And Second Derivatives Of Functions

Example Question #46 : First And Second Derivatives Of Functions

Example Question #47 : First And Second Derivatives Of Functions

All Calculus 2 Resources

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