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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #1 : Derivative At A Point

Find the derivative of the following function at the point $x=\frac{\pi}2$ .

f(x)=sin(x)ln(x)

Possible Answers:

$\frac{\pi}{2}$

0.45

$\frac{2}{\pi}$

Correct answer:

$\frac{2}{\pi}$

Explanation:

Here, we must use the product rule.

Assuming f(x)=sin(x) and g(x)= ln(x) , our expression becomes

$cos(x)ln(x)+\frac{sin(x)}{x}$ .

The question asks us to evaluate this at $x=\frac{\pi}2$ .

$cos\left(\frac{\pi}{2}\right)ln\left(\frac{\pi}{2}\right)+\frac{sin(\frac{\pi}{2})}{\frac{\pi}{2}}=\frac{2}{\pi}$

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Example Question #1 : Derivative At A Point

Given the function $f(x)=2x^{2}-4x+7$ , what is the slope at the point (5,10) ?

Possible Answers:

Correct answer:

Explanation:

As slope is defined as the derivative of a given function at a given point, we will need to take the derivative of $f(x)=2x^{2}-4x+7$ and substitute in the -value of the point (5,10) .

Using the Power Rule $\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ , $f'(x)=(2)2x^{2-1}-(1)4x^{1-1}+(0)7=4x-4$ . Subbing in x=5 , f'(5)=4(5)-4=20-4=16 .

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Example Question #91 : Derivative Review

Given the function $f(x)=5x^{2}+9x-12$ , what is the slope at the point (-1,3) ?

Possible Answers:

Correct answer:

Explanation:

As slope is defined as the derivative of a given function at a given point, we will need to take the derivative of $f(x)=5x^{2}+9x-12$ and substitute in the -value of the point (-1,3) .

Using the Power Rule $\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ , $f'(x)=(2)5x^{2-1}+(1)9x^{1-1}-(0)12=10x+9$ , Subbing in x=-1 , we get f'(-1)=10(-1)+9=-10+9=-1 .

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Example Question #92 : Derivative Review

Given the function $f(x)=x^{2}-7x-6$ , what is the slope at the point (2,9) ?

Possible Answers:

Correct answer:

Explanation:

As slope is defined as the derivative of a given function at a given point, we will need to take the derivative of $f(x)=x^{2}-7x-6$ and substitute in the -value of the point (2,9) .

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

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

Swapping in x=2 , we get f'(2)=2(2)-7=4-7=-3 .

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Example Question #93 : Derivative Review

Given $f(x)=7x^{2}-2x+9$ , find the value of f'(x) at the point (2,5) .

Possible Answers:

Correct answer:

Explanation:

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

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

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

Plugging in the -value of the point (2,5) into f'(x) , we get

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

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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 #91 : Derivatives

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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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1 : Derivative At A Point

Example Question #1 : Derivative At A Point

Example Question #91 : Derivative Review

Example Question #92 : Derivative Review

Example Question #93 : Derivative Review

Example Question #94 : Derivatives

Example Question #95 : Derivatives

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

Example Question #91 : Derivatives

Example Question #91 : Derivatives

All Calculus 2 Resources

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