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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 #41 : Derivative At A Point

Find the slope of the tangent line to the function $h(x)=\frac{2ln(x)}{x^2}$ at x=1 .

Possible Answers:

ln(2)-1

Correct answer:

Explanation:

The slope of the tangent line to a function at a point is the value of the derivative of the function at that point. In this problem, h(x) is a quotient of two functions, $2ln(x) \ and\ x^2$ , so the quotient rule is needed.

In general, the quotient rule is

$\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}$ .

To apply the quotient rule in this example, you must also know that $\frac{d}{dx}ln(x)=\frac{1}{x}$ and that $\frac{d}{dx}x^n=nx^{n-1}$ .

Therefore, the derivative is

$h'(x)=\frac{x^2(\frac{2}{x})-2ln(x)(2x)}{(x^2)^2}=\frac{2x-4xln(x)}{x^4}$

The last step is to substitute for in the derivative, which will tell us the slope of the tangent line to h(x) at x = 1 .

$h'(1)=\frac{2(1)-4(1)ln(1)}{1^4}=\frac{2-0}{1}=2$

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Example Question #125 : Derivative Review

What is the slope of $f(x)=x^{2}-3x+7$ at the point (9,0) ?

Possible Answers:

Correct answer:

Explanation:

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

Since we have $f(x)=x^{2}-3x+7$ , we can use the Power Rule

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

$f'(x)=(2)x^{2-1}-(1)3x^{1-1}+(0)7=2x-3$ .

We also have a point (9,0) with a -coordinate , so the slope

f'(9)=2(9)-3=18-3=15 .

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Example Question #126 : Derivative Review

What is the slope of $f(x)=\frac{x^{2}}{2}+12x-5$ at the point (-3,3) ?

Possible Answers:

Correct answer:

Explanation:

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

Since we have $f(x)=\frac{x^{2}}{2}+12x-5$ , we can use the Power Rule

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

$f'(x)=2\left(\frac{x^{2-1}}{2}\right)+(1)12x^{1-1}-(0)5=x+12$ .

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

f'(-3)=(-3)+12=9 .

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Example Question #127 : Derivative Review

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

Possible Answers:

Correct answer:

Explanation:

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

Since we have $f(x)=3x^{2}-2x+1$ , we can use the Power Rule

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

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

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

f'(1)=6(1)-2=6-2=4 .

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Example Question #121 : Derivatives

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

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

Possible Answers:

Correct answer:

Explanation:

The derivative of the function is

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

and was found using the following rules:

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

where f(x)=2x , $g(x)=\cos^2(x)$ , and $\frac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)$

Simply plug in x=0 into the first derivative function and solve:

f'(0)=2+0=2

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Example Question #131 : Derivative Review

What is the slope of a function $f(x)=3x^{2}+7x-5$ at (2,6) ?

Possible Answers:

Correct answer:

Explanation:

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

Since we have

$f(x)=3x^{2}+7x-5$ , we can use the Power Rule

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

$f'(x)=(2)3x^{2-1}+(1)7x^{1-1}-(0)5=6x+7$ .

We also have a point (2,6) with a -coordinate , so the slope

f'(2)=6(2)+7=12+7=19 .

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Example Question #131 : Derivatives

What is the slope of a function $f(x)=5x^{2}-9x+14$ at (-4,2) ?

Possible Answers:

Correct answer:

Explanation:

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

Since we have

$f(x)=5x^{2}-9x+14$ , we can use the Power Rule

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

$f'(x)=(2)5x^{2-1}-(1)9x^{1-1}+(0)14=10x-9$ .

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

f'(-4)=10(-4)-9=-40-9=-49 .

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Example Question #131 : Derivative Review

What is the slope of a function $f(x)=2x^{2}+13x-7$ at (-1,1) ?

Possible Answers:

Correct answer:

Explanation:

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

Since we have

$f(x)=2x^{2}+13x-7$ , we can use the Power Rule

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

$f'(x)=(2)2x^{2-1}+(1)13x^{1-1}-(0)7=4x+13$ .

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

f'(-1)=4(-1)+13=-4+13=9 .

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Example Question #11 : Derivative Defined As Limit Of Difference Quotient

Find f'(0) for

$f(x)=x^3\ln(x+1)$

Possible Answers:

$\infty$

Correct answer:

Explanation:

In order to find the derivative, we need to find f'(x) . We can find this by remembering the product rule and knowing the derivative of natural log.

Product Rule:

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

Derivative of natural log:

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

$f'(x)=\frac{g'(x)}{g(x)}$

Now lets apply this to our problem.

$f(x)=x^3\ln(x+1)$

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

$f'(0)=3(0)^2\ln(0+1)+\frac{0^3}{0+1}=0$

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Example Question #51 : Derivative At A Point

Find the derivative of the following function at the point z=0 :
$f(z)=\frac{z^2\cos(z)}{2}$

Possible Answers:

Correct answer:

Explanation:

The derivative of the function is

$f'(z)=\frac{4z\cos(z)-2z^2\sin(z)}{4}$

which was found using the following rules:

$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}$ ,

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

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

Finally, plug in the point z=0 into the first derivative function:

$f'(0)=\frac{0}{4}=0$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #41 : Derivative At A Point

Example Question #125 : Derivative Review

Example Question #126 : Derivative Review

Example Question #127 : Derivative Review

Example Question #121 : Derivatives

Example Question #131 : Derivative Review

Example Question #131 : Derivatives

Example Question #131 : Derivative Review

Example Question #11 : Derivative Defined As Limit Of Difference Quotient

Example Question #51 : Derivative At A Point

All Calculus 2 Resources

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