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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #101 : Derivative Review

Given $f(x)=5x^{2}-7x+2$ , what is the value of the slope at the point (-3,-1) ?

Possible Answers:

Correct answer:

Explanation:

Slope is defined as the derivative of a function at a given point.

By the Power Rule,

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

$f'(x)=(2)5x^{2-1}-(1)7x^{1-1}+(0)2=10x-7$ .

At (-3,-1) the -value is x=-3 , so the slope

f'(-3)=10(-3)-7=(-30)-7=-37 .

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Example Question #101 : Derivatives

Given $f(x)=2x^{2}-9x+5$ , what is the value of the slope at the point (4,-1) ?

Possible Answers:

Correct answer:

Explanation:

Slope is defined as the derivative of a function at a given point.

By the Power Rule,

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

$f'(x)=(2)2x^{2-1}-(1)9x^{1-1}+0(5)=4x-9$ .

At (4,-1) the -value is x=4 , so the slope

f'(4)=4(4)-9=16-9=7 .

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Example Question #51 : Derivatives

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

$h(x)=2x\sin(x)$

Possible Answers:

$2\cos(2)$

$2\sin(2)$

$2\sin(2)-4\cos(2)$

$2\sin(2)+4\cos(2)$

Correct answer:

$2\sin(2)+4\cos(2)$

Explanation:

The derivative of the function is given by the product rule:

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

Simply find the derivative of each function:

$\frac{d}{dx}(2x)=2$

$\frac{d}{dx}\sin(x)=\cos(x)$

The derivatives were found using the following rules:

$\frac{d}{dx}\sin(x)=\cos(x)$ , $\frac{d}{dx}(x^n)=nx^{n-1}$

Simply evaluate each derivative and the original functions at the point given, using the above product rule.

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Example Question #1222 : Calculus Ii

Find the derivative of the following function about the point (0, 1) :

$f(x)=3x^2+\cos(x)$

Possible Answers:

Correct answer:

Explanation:

The derivative of the function is

$f'(x)=6x-\sin(x)$

and was found using the following rules:

$\frac{d}{dx}(x^n)=nx^{n-1}$ , $\frac{d}{dx}\cos(x)= -\sin(x)$

Next, plug in the point we were asked to find the derivative at to finish the problem:

$6(0)-\sin(0)=0$

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Example Question #22 : Derivative At A Point

What is the slope of a function $f(x)=2x^{2}-5x+18$ at the point (3,4) ?

Possible Answers:

Correct answer:

Explanation:

By definition, slope is the first derivative of a given function f(x) .

Since $f(x)=2x^{2}-5x+18$ here, we can use the Power Rule

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

$f'(x)=(2)2x^{2-1}-(1)5x^{1-1}+(0)18=4x-5$ .

At (3,4) , x=3 and therefore the slope

f'(3)=4(3)-5=12-5=7 .

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Example Question #106 : Derivative Review

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

Possible Answers:

Correct answer:

Explanation:

By definition, slope is the first derivative of a given function f(x) .

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

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

$f'(x)=(2)4x^{2-1}+(1)x^{1-1}-(0)11=8x+1$ .

At the point (-2,5) , we have x=-2 , and so

f'(-2)=8(-2)+1=-16+1=-15 .

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Example Question #102 : Derivative Review

What is the slope of a function $f(x)=7x^{2}-2x+10$ at the point (-1,4) ?

Possible Answers:

None of the above

Correct answer:

Explanation:

By definition, slope is the first derivative of a given function f(x) .

Since $f(x)=7x^{2}-2x+10$ here, we can use the Power Rule

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

$f'(x)=(2)7x^{2-1}-(1)2x^{1-1}+(0)10=14x-2$ .

At the point (-1,4) , we have x=-1 , and so

f'(-1)=14(-1)-2=-14-2=-16 .

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Example Question #108 : Derivative Review

Given a function $f(x)=9x^{2}-12x+20$ , what is its slope at the point (4,5) ?

Possible Answers:

Correct answer:

Explanation:

Slope is defined as the first derivative of a function at a given point.

Given

$f(x)=9x^{2}-12x+20$ . we can use the Power Rule

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

$f'(x)=(2)9x^{2-1}-(1)12x^{(1-1)}+(0)20=18x-12$ .

Since the -coordinate of (4,5) is , the slope

f'(4)=18(4)-12=72-12=60 .

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Example Question #101 : Derivatives

Given a function $f(x)=-3x^{2}+14x-23$ , what is its slope at the point (-2,-2) ?

Possible Answers:

Correct answer:

Explanation:

Slope is defined as the first derivative of a function at a given point.

Given

$f(x)=-3x^{2}+14x-23$ . we can use the Power Rule

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

$f'(x)=-3(2)x^{2-1}+(1)14x^{1-1}-(0)23=-6x+14$ .

Since the -coordinate of (-2,-2) is , the slope

f'(-2)=-6(-2)+14=12+14=26 .

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Example Question #101 : Derivatives

Given a function $f(x)=\frac{1}{x^{2}}-3x+2$ , what is its slope at the point (4,1) ?

Possible Answers:

$\frac{32}{97}$

$\frac{97}{32}$

$-\frac{97}{32}$

None of the above.

$-\frac{32}{97}$

Correct answer:

$-\frac{97}{32}$

Explanation:

Slope is defined as the first derivative of a function at a given point. Given $f(x)=\frac{1}{x^{2}}-3x+2$ or $f(x)=x^{-2}-3x+2$ . we can use the Power Rule ( $\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ ) to derive $f'(x)=(-2))x^{-2-1}-(1)3x^{1-1}+(0)2=-2x^{-3}-3=-\frac{2}{x^{3}}-3$ . Since the -coordinate of (4,1) is , the slope $f'(4)=-\frac{2}{4^{3}}-3=-\frac{2}{64}-3=-\frac{1}{32}-3=-\frac{1}{32}-\frac{96}{32}=-\frac{97}{32}$ .

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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #101 : Derivative Review

Example Question #101 : Derivatives

Example Question #51 : Derivatives

Example Question #1222 : Calculus Ii

Example Question #22 : Derivative At A Point

Example Question #106 : Derivative Review

Example Question #102 : Derivative Review

Example Question #108 : Derivative Review

Example Question #101 : Derivatives

Example Question #101 : Derivatives

All Calculus 2 Resources

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