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Calculus 2 : Derivative Review

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Example Questions

Example Question #21 : Definition Of Derivative

Using the limit definition of a derivative, find the velocity function of a particle if its position is given by:

s(t)=4t^2+t+3

Possible Answers:

8t+1

8t+3

Correct answer:

8t+1

Explanation:

The limit definition of a derivative is

$\lim_{h\rightarrow 0}\left[\frac{f(x+h)-f(x)}{h}\right]$

where h represents a very small change in x.

Because the velocity function is simply the first derivative of the position function, we can use the above formula, with the position function as f(x), to find the velocity function:

$\lim_{h\rightarrow 0}\frac{4(t+h)^2+(t+h)+3-(4t^2+t+3)}{h}$

which simplified becomes

$\lim_{h\rightarrow 0}\frac{h(8t+1)}{h}=8t+1=v(t)$

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Example Question #21 : Derivative Review

What is the derivative of $f(x)=x^{200}\sin x$ ?

Possible Answers:

$f'(x)=x^{199}(200\sin x+x\cos x)$

$f'(x)=-\frac{x^{201}}{201}\cos x$

$f'(x)=200x^{199}\cos x$

$f'(x)=x^{199}(200\cos x+x\sin x)$

Correct answer:

$f'(x)=x^{199}(200\sin x+x\cos x)$

Explanation:

The derivative of

$f(x)=x^{200}\sin x$

is found using the chain rule:

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

So we have

$f'(x)=200x^{199}\sin x+x^{200}\cos x=x^{199}(200\sin x+x\cos x)$

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Example Question #23 : Definition Of Derivative

What is the derivative of

$f(x)=(\sin x^2+x^5-x)^7$ ?

Possible Answers:

$f'(x)=7(2x\cos x^2+5x^4-1)(\sin x^2+x^5-x)^6$

$f'(x)=(2x\cos x^2+5x^4-1)(\sin x^2+x^5-x)^7$

$f'(x)=7(-2x\cos x^2+5x^4-1)(\sin x^2+x^5-x)^6$

$f'(x)=(2x\cos x^2+5x^4-1)(\sin x^2+x^5-x)^6$

Correct answer:

$f'(x)=7(2x\cos x^2+5x^4-1)(\sin x^2+x^5-x)^6$

Explanation:

To find the derivative of $f(x)=(\sin x^2+x^5-x)^7$ , we simply use the chain rule, which is

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

So then we have

$f'(x)=7(2x\cos x^2+5x^4-1)(\sin x^2+x^5-x)^6$

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Example Question #24 : Definition Of Derivative

Given f(x)=5x^2+2 , what is f'(10) ?

Possible Answers:

f'(10)=502

f'(10)=2

f'(10)=100

f'(10)=25

Correct answer:

f'(10)=100

Explanation:

We can find the derivative of f(x)=5x^2+2 and simply plug in x=10 to get f'(10) :

To find the derivative we will need to use the power rule which states,

$x^n\ dx\rightarrow nx^{n-1}$ .

Also recall that the derivative of a constant is zero.

Applying the power rule we get the following.

$f'(x)=2\cdot 5x^{2-1}$

f'(x)=10x

so then we get

$f'(10)=10\cdot 10=100$ .

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Example Question #25 : Definition Of Derivative

Given $f(x)=\sin x^2$ , what is $f'\left(\sqrt{\frac{\pi}{2}}\right)$ ?

Possible Answers:

$f'\left(\sqrt{\frac{\pi}{2}}\right)=\sqrt{\frac{\pi}{2}}$

$f'\left(\sqrt{\frac{\pi}{2}}\right)=0$

$f'\left(\sqrt{\frac{\pi}{2}}\right)=\sqrt{\pi}$

$f'\left(\sqrt{\frac{\pi}{2}}\right)=2\sqrt{\frac{\pi}{2}}$

Correct answer:

$f'\left(\sqrt{\frac{\pi}{2}}\right)=0$

Explanation:

We can find the derivative of $f(x)=\sin x^2$ and simply plug in $x=\sqrt{\frac{\pi}{2}}$ to get $f'\left(\sqrt{\frac{\pi}{2}}\right)$ :

To find the derivative of this function we will need to use the chain rule which states,

$f(g(x))\ dx\rightarrow f'(g(x))\cdot g'(x)$

the power rule which states,

$x^n\ dx\rightarrow nx^{n-1}$ .

Also recall that the derivative of sine is cosine.

Applying these rules we get the following derivative.

$f'(x)=\cos x^2\cdot 2x^{2-1}$

$f'(x)=2x\cos x^2$

so then we get

$f'\left(\sqrt{\frac{\pi}{2}}\right)=2\sqrt{\frac{\pi}{2}}\cos \frac{\pi}{2}=2\sqrt{\frac{\pi}{2}}\cdot 0=0$

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Example Question #26 : Definition Of Derivative

Given f(x)=x^3+x^2+1 , what is f'(1) ?

Possible Answers:

f'(1)=5

f'(1)=0

f'(1)=1

f'(1)=6

Correct answer:

f'(1)=5

Explanation:

We can find the derivative of f(x)=x^3+x^2+1 and simply plug in x=1 to get f'(1) :

To find the derivative of this function we will need to use the power rule which states,

$x^n \ dx\rightarrow nx^{n-1}$ .

Applying the power rule we get the following.

$f'(x)=3x^{3-1}+2x^{2-1}+0$

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

so then we get

$f'(1)=3\cdot 1^2+2\cdot 1=5$ .

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Example Question #1151 : Calculus Ii

Given $f(x)=e^{x^2+1}$ , what is f'(5) ?

Possible Answers:

$f'(5)=e^{26}$

$f'(5)=5e^{26}$

$f'(5)=10e^{13}$

$f'(5)=10e^{26}$

Correct answer:

$f'(5)=10e^{26}$

Explanation:

We can find the derivative of $f(x)=e^{x^2+1}$ and simply plug in x=5 to get f'(5) :

To find the derivative of this function we will need to use the power rule, chain rule, and rule of exponentials.

Power Rule:

$x^n \ dx=nx^{n-1}$

Chain Rule:

$f(g(x)) \ dx=f'(g(x))\cdot g'(x)$

Rule of Exponentials:

$e^u \ dx=e^u \frac{du}{dx}$

Applying these rules we get the following.

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

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

$f'(x)=2xe^{x^2+1}$

so then we get

$f'(5)=10e^{25+1}=10e^{26}$ .

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Example Question #28 : Definition Of Derivative

What is the derivative of $f(x)=e^{x^3}$ ?

Possible Answers:

$f'(x) = 3xe^{x^3}$

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

$f'(x) = 3e^{x^3}$

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

Correct answer:

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

Explanation:

With a simple application of the chain rule, which is

f(x)=g(h(x))

$\Rightarrow f'(x)=g'(h(x))h'(x)$

with g(x)=e^x and h(x)=x^3 we get that the derivative of $f(x)=e^{x^3}$ is

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

since h'(x)=3x^2 and g'(x)=e^x .

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Example Question #29 : Definition Of Derivative

What is the derivative of the function $f(x)=e^{x^2+x}$ ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

With a simple application of the chain rule, which is

f(x)=g(h(x))

$\Rightarrow f'(x)=g'(h(x))h'(x)$

we get that the derivative of $f(x)=e^{x^2+x}$ is

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

since h'(x)=2x+1 .

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Example Question #30 : Definition Of Derivative

What is the derivative of the function

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

Possible Answers:

$f'(x) = 2xe^{\sin x^2}$

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

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

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

Correct answer:

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

Explanation:

With an application of the chain rule, which is

f(x)=g(h(x))

$\Rightarrow f'(x)=g'(h(x))h'(x)$

we get that the derivative of $f(x)=e^{\cos x^2}$ is

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

Because we have that g(x)=e^x and $h(x)=\cos x^2$ which means

g'(x)=e^x

$h'(x)=-2x\sin x^2$ .

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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #21 : Definition Of Derivative

Example Question #21 : Derivative Review

Example Question #23 : Definition Of Derivative

Example Question #24 : Definition Of Derivative

Example Question #25 : Definition Of Derivative

Example Question #26 : Definition Of Derivative

Example Question #1151 : Calculus Ii

Example Question #28 : Definition Of Derivative

Example Question #29 : Definition Of Derivative

Example Question #30 : Definition Of Derivative

All Calculus 2 Resources

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