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Calculus 2 : Derivative Review

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Example Questions

Example Question #61 : Derivative At A Point

What is the slope of $f(x)=x^{2}+4x+5$ at (2,-2) ?

Possible Answers:

Correct answer:

Explanation:

We define slope as the first derivative of a given function.

Since we have $f(x)=x^{2}+4x+5$ , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ to determine that

$f'(x)=(2)x^{2-1}+(1)4x^{1-1}+(0)5=2x+4$ .

We also have a point (2,-2) with a -coordinate , so the slope

f'(2)=2(2)+4=4+4=8 .

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Example Question #62 : Derivative At A Point

What is the slope of $f(x)=3x^{2}-5x+6$ at (5,-3) ?

Possible Answers:

Correct answer:

Explanation:

We define slope as the first derivative of a given function.

Since we have $f(x)=3x^{2}-5x+6$ , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ to determine that

$f'(x)=(2)3x^{2-1}-(1)5x^{1-1}+(0)6=6x-5$ .

We also have a point (5,-3) with a -coordinate , so the slope

f'(5)=6(5)-5=30-5=25 .

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Example Question #71 : Derivative At A Point

Find f'(0) for

$f(x)=\sin(x)\ln(x+1)$

Possible Answers:

$\pi$

$\frac{\pi}{4}$

$\frac{\pi}{2}$

Correct answer:

Explanation:

In order to take the derivative, we need to remember how to take derivative of natural log and the product rule.

Derivative of natural log:

$f(t)=\ln(g(t))$

$f'(t)=\frac{g'(t)}{g(t)}$

Product Rule:

[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x)

Now lets apply both of these rule to our problem.

$f(x)=\sin(x)\ln(x+1)$

$f'(x)=\cos(x)\ln(x+1)+\frac{\sin(x)}{x+1}$

We plug in 0 to get.

$f'(0)=\cos(0)\ln(1)+\frac{\sin(0)}{1}$

$f'(0)=1*0+\frac{0}{1}=0$

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Example Question #72 : Derivative At A Point

Find f'(1) for

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

Possible Answers:

$2\cos(1)$

The derivative doesn't exist.

$2\cos(1)+\sin(1)$

$2\tan(1)$

$2\cos(1)-\sin(1)$

Correct answer:

$2\cos(1)-\sin(1)$

Explanation:

In order to find f'(1) , we first find f'(x) .

To find the derivative, we need to remember the product rule.

Product Rule:

[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x) .

Lets apply this rule to our problem.

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

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

We plug in 1 to get

$f'(1)=2(1)\cos(1)-(1)^2\sin(1)$

$=2\cos(1)-\sin(1)$

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Example Question #73 : Derivative At A Point

Find f'(0) for

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

Possible Answers:

$\log(2)$

$8\ln(2)$

Correct answer:

$8\ln(2)$

Explanation:

In order to find f'(0) , we need to find f'(x) .

We need to remember the derivative of expotential functions.

Derivative of expotential functions:

$f(x)=a^{g(x)}$

$f'(x)=g'(x)a^{g(x)}\ln(a)$

Now lets apply this to our problem.

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

$f'(x)=2*2^{2x+2}*\ln(2)$

$f'(0)=2*2^{2(0)+2}*\ln(2)=8\ln(2)$

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Example Question #74 : Derivative At A Point

Find f'(0) for

$f(t)=t^3+2^{26t^2}$

Possible Answers:

$\log(2)$

Correct answer:

Explanation:

In order to take the derivative, we need to remember how to take derivatives of exponential functions not in base e.

Derivatives of exponential functions not in base e:

$f(x)=a^{g(x)}$

$f'(x)=g'(x)a^{g(x)}\ln(a)$

Now lets apply this to our problem.

$f(t)=t^3+2^{26t^2}$

$f'(t)=3t^2+52t\ 2^{26t^2}\ln(2)$

Now we plug in 0 to get.

$f'(0)=3(0)^2+52(0) \2^{26(0)^2}\ln(2)$

$f'(0)=0+0*1*\log(2)=0$

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Example Question #157 : Derivative Review

What is the slope of $f(x)=-x^{2}+10x-5$ at (-5,3) ?

Possible Answers:

Correct answer:

Explanation:

We define slope as the first derivative of a given function.

Since we have $f(x)=-x^{2}+10x-5$ , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ to determine that

$f'(x)=(2)(-1)x^{2-1}+(1)10x^{1-1}-(0)5=-2x+10$ .

We also have a point (-5,3) with a -coordinate , so the slope

f'(-5)=-2(-5)+10=10+10=20 .

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Example Question #75 : Derivative At A Point

What is the slope of $f(x)=7x^{2}-2x+9$ at (2,0) ?

Possible Answers:

Correct answer:

Explanation:

We define slope as the first derivative of a given function.

Since we have

$f(x)=7x^{2}-2x+9$ , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$

to determine that

$f'(x)=(2)7x^{2-1}-(1)2x^{1-1}+(0)9=14x-2$ .

We also have a point (2,0) with a -coordinate , so the slope

f'(2)=14(2)-2=28-2=26 .

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Example Question #159 : Derivative Review

What is the slope of $f(x)=6x^{2}+7x+1$ at (-1,4) ?

Possible Answers:

Correct answer:

Explanation:

We define slope as the first derivative of a given function.

Since we have

$f(x)=6x^{2}+7x+1$ , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$

to determine that

$f'(x)=(2)6x^{2-1}+(1)7x^{1-1}+(0)1=12x+7$ .

We also have a point (-1,4) with a -coordinate , so the slope

f'(-1)=12(-1)+7=-12+7=-5 .

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Example Question #76 : Derivative At A Point

Find the derivative of the following function at x=0 :

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

Possible Answers:

4e^1

-2e^1

2e^1

Correct answer:

2e^1

Explanation:

The derivative of the function is

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

and was found using the following rules:

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

Now, just plug zero into the first derivative function to get our answer:

f'(0)=2e^1

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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #61 : Derivative At A Point

Example Question #62 : Derivative At A Point

Example Question #71 : Derivative At A Point

Example Question #72 : Derivative At A Point

Example Question #73 : Derivative At A Point

Example Question #74 : Derivative At A Point

Example Question #157 : Derivative Review

Example Question #75 : Derivative At A Point

Example Question #159 : Derivative Review

Example Question #76 : Derivative At A Point

All Calculus 2 Resources

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