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Calculus 2 : Derivatives

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #111 : Derivative Review

Find the derivative of $y=\left | x-3\right |$ at x=3 .

Possible Answers:

$\pm1$

Does not exist.

Correct answer:

Does not exist.

Explanation:

Split the absolute value into both positive and negative components.

$y=\left | x-3\right | = \pm(x-3)$

Take their derivatives.

$\frac{d}{dx}[y=x-3]=1$

$\frac{d}{dx}[y=-(x-3)]=-1$

At x=3 , there exists a spike in the graph. For spikes, the derivative does not exist under this exception.

The answer is:

$\textup{Does not exist.}$

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Example Question #111 : Derivative Review

What is the slope of a function $f(x)=10x^{2}-25x+12$ at the point (5,7) ?

Possible Answers:

100

Correct answer:

Explanation:

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

We are given the function

$f(x)=10x^{2}-25x+12$ and a point (5,7) , so we need to find the derivative f'(x) and solve for the point's -coordinate.

Using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all nonzero , we can derive

$f'(x)=(2)10x^{2-1}-(1)25x^{1-1}+(0)12=20x-25$ .

Substituting the -coordinate , we have a slope:

f'(5)=20(5)-25=100-25=75 .

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Example Question #31 : Derivative At A Point

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

Possible Answers:

$\frac{1}{3}$

None of the above

Correct answer:

Explanation:

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

We are given the function

$f(x)=x^{2}+7x-3$ and a point (-3,0) , so we need to find the derivative f'(x) and solve for the point's -coordinate.

Using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all nonzero , we can derive

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

Substituting the -coordinate , we have a slope:

f'(-3)=2(-3)+7=-6+7=1 .

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Example Question #111 : Derivatives

What is the slope of a function $f(x)=6x^{2}-12x+9$ at the point (-10,10) ?

Possible Answers:

132

None of the above

Correct answer:

Explanation:

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

We are given the function

$f(x)=6x^{2}-12x+9$ and a point (-10,10) , so we need to find the derivative f'(x) and solve for the point's -coordinate.

Using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all nonzero , we can derive

$f'(x)=(2)6x^{2-1}-(1)12x^{1-1}+(0)9=12x-12$ .

Substituting the -coordinate , we have a slope:

f'(-10)=12(-10)-12=-120-12=-132 .

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Example Question #112 : Derivatives

What is the slope of $f(x)=4x^{2}-5x+12$ at (1,-1) ?

Possible Answers:

Correct answer:

Explanation:

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

Since,

$f(x)=4x^{2}-5x+12$ , we can use the Power Rule

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

$f'(x)=(2)4x^{2-1}-(1)5x^{1-1}+(0)12=8x-5$ .

At the point (1,-1) , the -coordinate is .

Thus, the slope is

f'(1)=8(1)-5=8-5=3 .

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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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Calculus 2 : Derivatives

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #111 : Derivative Review

Example Question #111 : Derivative Review

Example Question #31 : Derivative At A Point

Example Question #111 : Derivatives

Example Question #112 : Derivatives

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

All Calculus 2 Resources

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