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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #94 : Derivatives

Given $f(x)=5x^{2}+6x-12$ , find the value of f'(x) at the point (-1,-1) .

Possible Answers:

Correct answer:

Explanation:

Given the function $f(x)=5x^{2}+6x-12$ , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ to find its derivative:

$f'(x)=(2)5x^{2-1}+(1)6x^{1-1}-(0)12=10x+6$ .

Plugging in the -value of the point (-1,-1) into f'(x) , we get

f'(-1)=10(-1)+6=-10+6=-4 .

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Example Question #95 : Derivatives

Given $f(x)=10x^{2}-9x+1$ , find the value of f'(x) at the point (-2,7) .

Possible Answers:

Correct answer:

Explanation:

Given the function $f(x)=10x^{2}-9x+1$ , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ to find its derivative:

$f'(x)=(2)10x^{2-1}-(1)9x^{1-1}+(0)=20x-9$ .

Plugging in the -value of the point (-2,7) into f'(x) , we get

f'(-2)=20(-2)-9=-40-9=-49 .

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Example Question #12 : Derivative Defined As Limit Of Difference Quotient

Find the derivative of y=(2x+4)(3x+2) at point (1,30) .

Possible Answers:

Correct answer:

Explanation:

Use either the FOIL method to simplify before taking the derivative or use the product rule to find the derivative of the function.

The product rule will be used for simplicity.

y'=(2x+4)(3)+(3x+2)(2) = 6x+12 + 6x+4= 12x+16

Substitute x=1 .

y'(1)= 12(1)+16=28

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Example Question #91 : Derivatives

Given the function $\small f(x)=2(x^2-1)^6$ , calculate $\small f'(2)$ .

Possible Answers:

$\small \small \small \small f'(2)=11,664$

$\small \small \small \small \small f'(2)=11,663$

$\small \small \small \small \small f'(2)=1458$

$\small \small \small \small \small f'(2)=2,916$

Correct answer:

$\small \small \small \small f'(2)=11,664$

Explanation:

The derivative of $\small f(x)=2(x^2-1)^6$ can be computed using the chain rule:

$\small \small f'(x)=2\cdot 6\cdot 2x(x^2-1)^5=24x(x^2-1)^5$

so now we just plug in $\small x=2$ :

$\small \small \small f'(2)=24\cdot2(2^2-1)^5=48\cdot 3^5=11,664$

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Example Question #95 : Derivative Review

Given $f(x)=x^{2}-7x+6$ , what is the value of the slope at the point (2,3) ?

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 $n\neq0$ ,

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

At (2,3) the -value is x=2 , so the slope

f'(2)=2(2)-7=4-7=-3 .

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Example Question #101 : Derivatives

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 #102 : Derivative Review

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:

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

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

$2\sin(2)$

$2\cos(2)$

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 #21 : Derivative At A Point

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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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #94 : Derivatives

Example Question #95 : Derivatives

Example Question #12 : Derivative Defined As Limit Of Difference Quotient

Example Question #91 : Derivatives

Example Question #95 : Derivative Review

Example Question #101 : Derivatives

Example Question #102 : Derivative Review

Example Question #51 : Derivatives

Example Question #21 : Derivative At A Point

Example Question #22 : Derivative At A Point

All Calculus 2 Resources

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