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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 #111 : Derivatives

What is the slope of $f(x)=x^{2}+2x+11$ at (3,0) ?

Possible Answers:

Correct answer:

Explanation:

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

Since,

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

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

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

At the point (3,0) , the -coordinate is .

Thus, the slope is

f'(3)=2(3)+2=6+2=8 .

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Example Question #114 : Derivatives

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

Possible Answers:

Correct answer:

Explanation:

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

Since,

$f(x)=7x^{2}-6$ , we can use the Power Rule

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

$f'(x)=(2)7x^{1-1}-(0)6=14x$ .

At the point (5,3) , the -coordinate is .

Thus, the slope is f'(5)=14(5)=70 .

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Example Question #1241 : Calculus Ii

What is the slope of a function f(x)=5x^2-x+7 at the point (4,5) ?

Possible Answers:

None of the above

Correct answer:

Explanation:

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

Since f(x)=5x^2-x+7 , we can use the Power Rule

$\frac{d}{dx}x^n=nx^{n-1}$ for all $n\neq 0$ to determine that

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

Since we're given a point (4,5) , we can use the -coordinate to solve for the slope at that point.

Thus,

f'(4)=10(4)-1=40-1=39 .

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Example Question #13 : Derivative Defined As Limit Of Difference Quotient

What is the slope of a function f(x)=-2x^2+5x-3 at the point (0,1) ?

Possible Answers:

Correct answer:

Explanation:

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

Since f(x)=-2x^2+5x-3 , we can use the Power Rule

$\frac{d}{dx}x^n=nx^{n-1}$ for all $n\neq 0$ to determine that

$f'(x)=(2)(-2)x^{2-1}+(1)5x^{1-1}-(0)3=-4x+5$ .

Since we're given a point (0,1) , we can use the x-coordinate to solve for the slope at that point.

Thus,

f'(0)=-4(0)+5=5.

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Example Question #116 : Derivatives

What is the slope of a function f(x)=-x^2+8x-4 at the point (-1,0) ?

Possible Answers:

$-\frac{1}{10}$

$\frac{1}{10}$

None of the above

Correct answer:

Explanation:

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

Since f(x)=-x^2+8x-4 , we can use the Power Rule

$\frac{d}{dx}x^n=nx^{n-1}$ for all $n\neq 0$ to determine that

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

Since we're given a point (-1,0) , we can use the x-coordinate to solve for the slope at that point.

Thus,

f'(-1)=-2(-1)+8=2+8=10 .

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Example Question #1241 : Calculus Ii

What is the slope of the tangent line to the function

$h(x)=e^{x}ln(x)$

when x = 1?

Possible Answers:

$\frac{1}{e}$

Correct answer:

Explanation:

The slope of the tangent line to a function at a point is the value of the derivative at that point. To calculate the derivative in this problem, the product rule is necessary. Recall that the product rule states that:

$\frac{d}{dx}(m(x)n(x))=m'(x)n(x)+m(x)n'(x)$ .

In this example, $m(x)=e^x \ and\ n(x)=ln(x).$

Therefore,

$m'(x)=e^x \ and\ n'(x)=\frac{1}{x}$ , and

$\frac{d}{dx}(e^xln(x))=e^xln(x)+e^x\cdot\frac{1}{x}$

At x = 1, this dervative has the value

$e^1ln(1)+e^1\cdot\frac{1}{1}=0+e=e$ .

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Example Question #32 : Derivative At A Point

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

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

Possible Answers:

Correct answer:

Explanation:

The derivative of the function is

$f'(x)=-2\cos(x)\sin(x)+3$

and was found using the following rules:

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

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

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

where $f(x)=\cos^2(x), g(x)=x$ in the chain rule.

Plug in 0 in the derivative function to get f'(x)=3

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Example Question #1241 : Calculus Ii

What is the slope of $f(x)=x^{2}-6x+15$ at (2,-4) ?

Possible Answers:

Correct answer:

Explanation:

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

Since we have

$f(x)=x^{2}-6x+15$ , we can use the Power Rule

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

$f'(x)=(2)x^{2-1}-(1)6x^{1-1}+(0)15=2x-6$ .

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

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

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Example Question #123 : Derivative Review

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

Possible Answers:

None of the above

Correct answer:

Explanation:

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

Since we have

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

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

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

We also have a point (1,2) with a -coordinate , so the slope

f'(1)=8(1)+1=8+1=9 .

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Example Question #124 : Derivative Review

What is the slope of $f(x)=-x^{2}-5x-7$ at (4,-4) ?

Possible Answers:

None of the above

Correct answer:

Explanation:

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

Since we have

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

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

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

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

f'(4)=-2(4)-5=-8-5=-13 .

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #111 : Derivatives

Example Question #114 : Derivatives

Example Question #1241 : Calculus Ii

Example Question #13 : Derivative Defined As Limit Of Difference Quotient

Example Question #116 : Derivatives

Example Question #1241 : Calculus Ii

Example Question #32 : Derivative At A Point

Example Question #1241 : Calculus Ii

Example Question #123 : Derivative Review

Example Question #124 : Derivative Review

All Calculus 2 Resources

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