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Calculus 2 : Derivative Review

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Example Questions

Example Question #161 : Derivatives

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

$f(x)=5x^4e^{\sin(x)}$

Possible Answers:

Correct answer:

Explanation:

The derivative of the function is

$f'(x)=20x^3e^{\sin(x)}+5x^4\cos(x)e^{\sin(x)}$

and was found using the following rules:

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

To finish, plug in the point given into the first derivative function:

f'(0)=0+0=0

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Example Question #72 : Derivative At A Point

Find the derivative of the following function at x=2 :

$f(x)=x^2+\csc(2x)+e^\ln(x)$

Possible Answers:

$5+2\csc(x)\cot(x)$

$5-2\csc(4)\cot(4)$

$2x+\csc(2x)+\cot(2x)$

$6+\csc(2x)$

Correct answer:

$5-2\csc(4)\cot(4)$

Explanation:

The derivative of the function is:

$f'(x)=2x-2\csc(2x)\cot(2x)+1$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x}\csc(x)=-\csc(x)\cot(x)$ , $e^{\ln(x)}=x$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$

Now, just plug in the point x=2 into the first derivative function:

$f'(2)=5-2\csc(4)\cot(4)$

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Example Question #161 : Derivatives

What is the slope of $f(x)=2x^{2}-7x+11$ at (5,6) ?

Possible Answers:

Correct answer:

Explanation:

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

Since we have

$f(x)=2x^{2}-7x+11$ , we can use the Power Rule

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

to determine that

$f'(x)=(2)2x^{2-1}-(1)7x^{1-1}+(0)11=4x-7$ .

We also have a point (5,6) with a -coordinate , so the slope

f'(5)=4(5)-7=20-7=13 .

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Example Question #81 : Derivative At A Point

What is the slope of $f(x)=-5x^{2}+\frac{1}{2}x+4$ at (2,5) ?

Possible Answers:

$-\frac{2}{39}$

$\frac{2}{39}$

$\frac{39}{2}$

$-\frac{39}{2}$

None of the above

Correct answer:

$-\frac{39}{2}$

Explanation:

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

Since we have

$f(x)=-5x^{2}+\frac{1}{2}x+4$ , we can use the Power Rule

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

to determine that

$f'(x)=(2)(-5)x^{2-1}+(1)\frac{1}{2}x^{1-1}+(0)4=-10x+\frac{1}{2}$ .

We also have a point (2,5) with a -coordinate , so the slope

$f'(2)=-10(2)+\frac{1}{2}=-20+\frac{1}{2}=-\frac{40}{2}+\frac{1}{2}=-\frac{39}{2}$ .

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Example Question #82 : Derivative At A Point

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

Possible Answers:

Correct answer:

Explanation:

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

Since we have

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

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

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

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

f'(-1)=(-1)-1=-2 .

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Example Question #83 : Derivative At A Point

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

Possible Answers:

Correct answer:

Explanation:

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

Since we have

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

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

to determine that

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

We also have a point (2,-5) with a -coordinate , so the slope

f'(2)=-2(2)+5=-4+5=1 .

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Example Question #84 : Derivative At A Point

What is the slope of $f(x)=4x^{2}-3x+2$ at (1,-1) ?

Possible Answers:

Correct answer:

Explanation:

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

Since we have

$f(x)=4x^{2}-3x+2$ , we can use the Power Rule

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

$f'(x)=(2)4x^{2-1}-(1)3x^{1-1}+(0)2=8x-3$ .

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

f'(1)=8(1)-3=8-3=5 .

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Example Question #85 : Derivative At A Point

What is the slope of $f(x)=-x^{2}+11x+4$ at (4,-5) ?

Possible Answers:

Correct answer:

Explanation:

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

Since we have

$f(x)=-x^{2}+11x+4$ , we can use the Power Rule

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

to determine that

$f'(x)=(2)(-1)x^{2-1}+(1)11x^{1-1}+(0)4=-2x+11$ .

We also have a point (4,-5) with a -coordinate , so the slope

f'(4)=-2(4)+11=-8+11=3 .

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Example Question #81 : Derivative At A Point

Find f'(1) for

$f(x)=x^3+x^2+x+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}$

Possible Answers:

$\infty$

Correct answer:

Explanation:

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

$f(x)=x^3+x^2+x+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}$

$f'(x)=3x^2+2x+1-\frac{1}{x^2}-\frac{2}{x^3}-\frac{3}{x^4}$

Now we plug in 1 to get

$f'(1)=3(1)^2+2(1)+1-\frac{1}{(1)^2}-\frac{2}{(1)^3}-\frac{3}{(1)^4}$

f'(1)=3+2+1-1-2-3=0

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Example Question #87 : Derivative At A Point

What is the slope of $f(x)=3x^{2}-14$ at the point (4,-2) ?

Possible Answers:

Correct answer:

Explanation:

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

Since we have $f(x)=3x^{2}-14$ , we can use the Power Rule

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

to determine that

$f'(x)=(2)3x^{2-1}-(0)14=6x$ .

We also have a point (4,-2) with a -coordinate , so the slope

f'(4)=6(4)=24 .

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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #161 : Derivatives

Example Question #72 : Derivative At A Point

Example Question #161 : Derivatives

Example Question #81 : Derivative At A Point

Example Question #82 : Derivative At A Point

Example Question #83 : Derivative At A Point

Example Question #84 : Derivative At A Point

Example Question #85 : Derivative At A Point

Example Question #81 : Derivative At A Point

Example Question #87 : Derivative At A Point

All Calculus 2 Resources

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