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Calculus 2 : Derivative Review

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 #81 : Derivative At A Point

What is the slope of $f(x)=2x^{2}+11x-14$ at the point (-1,6) ?

Possible Answers:

$\frac{1}{7}$

$-\frac{1}{7}$

None of the above

Correct answer:

Explanation:

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

Since we have $f(x)=2x^{2}+11x-14$ , we can use the Power Rule

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

to determine that

$f(x)=(2)2x^{2-1}+(1)11x^{1-1}-(0)14=4x+11$ .

We also have a point (-1,6) with a -coordinate , so the slope

f(-1)=4(-1)+11=-4+11=7 .

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Example Question #172 : Derivative Review

What is the slope of $f(x)=\frac{1}{3}x^{2}-4x$ at the point (7,2) ?

Possible Answers:

$\frac{3}{2}$

$-\frac{2}{3}$

$-\frac{3}{2}$

$\frac{2}{3}$

None of the above

Correct answer:

$\frac{2}{3}$

Explanation:

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

Since we have $f(x)=\frac{1}{3}x^{2}-4x$ , we can use the Power Rule

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

to determine that

$f'(x)=(2)\frac{1}{3}x^{2-1}-(1)4x^{1-1}=\frac{2}{3}x-4$ .

We also have a point (7,2) with a -coordinate , so the slope

$f'(7)=\frac{2}{3}(7)-4=\frac{14}{3}-\frac{12}{3}=\frac{2}{3}$ .

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Example Question #171 : Derivative Review

Calculate the derivative of $f(x)=\frac{x^3}{3}-2x^2+x-3$ at the point x=3 .

Possible Answers:

Correct answer:

Explanation:

There are 2 steps to solving this problem.

First, take the derivative of f(x)

Then, replace the value of x with the given point and evaluate

For example, if x=a , then we are looking for the value of f'(x=a) , or the derivative of f(x) at x=a .

$f(x)=\frac{x^3}{3}-2x^2+x-3$

Calculate f'(x)

Derivative rules that will be needed here:

Derivative of a constant is 0. For example, $\frac{\mathrm{d} }{\mathrm{d} t} 5 = 0$
Taking a derivative on a term, or using the power rule, can be done by doing the following: $\frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}$

f'(x)=x^2 - 4x + 1

Then, plug in the value of x and evaluate

f'(x=3)=3^2-4*3+1=-2

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Example Question #1292 : Calculus Ii

Calculate the derivative of $f(x)=\frac{x^3}{12}-4x^2+9x-13$ at the point x=3 .

Possible Answers:

-12.75

-11.75

-13.75

-14.75

Correct answer:

-12.75

Explanation:

There are 2 steps to solving this problem.

First, take the derivative of f(x)

Then, replace the value of x with the given point and evaluate

For example, if x=a , then we are looking for the value of f'(x=a) , or the derivative of f(x) at x=a .

$f(x)=\frac{x^3}{12}-4x^2+9x-13$

Calculate f'(x)

Derivative rules that will be needed here:

Derivative of a constant is 0. For example, $\frac{\mathrm{d} }{\mathrm{d} t} 5 = 0$
Taking a derivative on a term, or using the power rule, can be done by doing the following: $\frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}$

$f'(x)=\frac{3x^2}{12}-8x+9$

Then, plug in the value of x and evaluate

$f'(x=3)=\frac{3^2}{4}-8*3+9= \frac{9}{4}-24+9=-12.75$

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Example Question #91 : Derivative At A Point

Calculate the derivative of $f(x)=\frac{9x^3}{16}+3x^2+2$ at the point x=4 .

Possible Answers:

Correct answer:

Explanation:

There are 2 steps to solving this problem.

First, take the derivative of f(x)

Then, replace the value of x with the given point and evaluate

For example, if x=a , then we are looking for the value of f'(x=a) , or the derivative of f(x) at x=a .

$f(x)=\frac{9x^3}{16}+3x^2+2$

Calculate f'(x)

Derivative rules that will be needed here:

Derivative of a constant is 0. For example, $\frac{\mathrm{d} }{\mathrm{d} t} 5 = 0$
Taking a derivative on a term, or using the power rule, can be done by doing the following: $\frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}$

$f'(x)=\frac{9*3*x^2}{16}+3*2*x$

Then, plug in the value of x and evaluate

$f'(x=4)=\frac{27*4^2}{16}+6*4=\frac{27*16}{16}+24=27+24=51$

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Example Question #1301 : Calculus Ii

Calculate the derivative of $f(x)=4e^{5x}+9x^2+5$ at the point x=0 .

Possible Answers:

Correct answer:

Explanation:

There are 2 steps to solving this problem.

First, take the derivative of f(x)

Then, replace the value of x with the given point and evaluate

For example, if x=a , then we are looking for the value of f'(x=a) , or the derivative of f(x) at x=a .

Derivative rules that will be needed here:

Derivative of a constant is 0. For example, $\frac{\mathrm{d} }{\mathrm{d} t} 5 = 0$
Taking a derivative on a term, or using the power rule, can be done by doing the following: $\frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}$
Special rule when differentiating an exponential function: $\frac{\mathrm{d} }{\mathrm{d} t} e^{kt} = ke^{kt}$ , where k is a constant.

$f(x)=4e^{5x}+9x^2+5$

Calculate f'(x) .

$f'(x)=20e^{5x}+18x$

Then, plug in the value of x and evaluate.

$f'(x=4)=20e^{5*0}+18*0=20*1+0=20$

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Example Question #1301 : Calculus Ii

Calculate the derivative of $f(x)=\frac{3e^{10x}}{5}+e^{11x}$ at the point x=0 .

Possible Answers:

Correct answer:

Explanation:

There are 2 steps to solving this problem.

First, take the derivative of f(x)

Then, replace the value of x with the given point and evaluate

For example, if x=a , then we are looking for the value of f'(x=a) , or the derivative of f(x) at x=a .

Derivative rules that will be needed here:

When differentiating an exponential function: $\frac{\mathrm{d} }{\mathrm{d} t} e^{kt} = ke^{kt}$ , where k is a constant.

$f(x)=\frac{3e^{10x}}{5}+e^{11x}$

Calculate f’'(x) .

$f'(x)=\frac{3*10e^{10x}}{5}+11e^{11x}$

Then, plug in the value of x and evaluate.

$f'(x=0)=\frac{3*10e^{10*0}}{5}+11e^{11*0} = 6*1+11*1=6+11=17$

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Example Question #1302 : Calculus Ii

Calculate the derivative of $f(x)=\frac{e^{4x}}{3} + e^{5x}$ at the point x=4 .

Possible Answers:

${4e^{16}}+5e^{20}$

${4e^{8}}+5e^{9}$

$\frac{4e^{8}}{3}+5e^{9}$

$\frac{4e^{16}}{3}+5e^{20}$

Correct answer:

$\frac{4e^{16}}{3}+5e^{20}$

Explanation:

There are 2 steps to solving this problem.

First, take the derivative of f(x)

Then, replace the value of x with the given point and evaluate

For example, if x=a , then we are looking for the value of f'(x=a) , or the derivative of f(x) at x=a .

$f(x)=\frac{e^{4x}}{3} + e^{5x}$

Calculate f'(x) .

Derivative rules that will be needed here:

When differentiating an exponential function: $\frac{\mathrm{d} }{\mathrm{d} t} e^{kt} = ke^{kt}$ , where k is a constant.

$f'(x)=\frac{4e^{4x}}{3}+5e^{5x}$

Then, plug in the value of x and evaluate.

$f'(x=4)=\frac{4e^{4*4}}{3}+5e^{5*4} = \frac{4e^{16}}{3}+5e^{20}$

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Example Question #1303 : Calculus Ii

Calculate the derivative of $f(x)=e^{2*x}+4x^3$ +x at the point x=10 .

Possible Answers:

$2e^{20}+1220$

$2e^{20}+1199$

$2e^{20}+1200$

$2e^{20}+1201$

Correct answer:

$2e^{20}+1201$

Explanation:

There are 2 steps to solving this problem.

First, take the derivative of f(x)

Then, replace the value of x with the given point and evaluate

For example, if x=a , then we are looking for the value of f'(x=a) , or the derivative of f(x) at x=a .

$f(x)=e^{2*x}+4x^3+x$

Calculate f'(x)

Derivative rules that will be needed here:

Taking a derivative on a term, or using the power rule, can be done by doing the following: $\frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}$
Special rule when differentiating an exponential function: $\frac{\mathrm{d} }{\mathrm{d} t} e^{kt} = ke^{kt}$ , where k is a constant.

$f'(x)=2e^{2x}+12x^2+1$

Then, plug in the value of x and evaluate.

$f'(x=10)=2e^{2*10}+12*10^2+1 = 2e^{20} + 12*100+1=2e^{20}+1201$

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Example Question #92 : Derivative At A Point

Calculate the derivative of f(x)=5sin(8x) at the point $x=\frac{\pi}{2}$ .

Possible Answers:

Correct answer:

Explanation:

There are 2 steps to solving this problem.

First, take the derivative of f(x)

Then, replace the value of x with the given point and evaluate

For example, if x=a , then we are looking for the value of f'(x) , or the derivative of f(x) at x=a .

f(x)=5sin(8x)

Calculate f'(x)

Derivative rules that will be needed here:

$\frac{\mathrm{d} }{\mathrm{d} t} sin(kt) = kcos(kt)$

f'(x)=5*8*cos(8x)

Then, plug in the value of x and evaluate.

$f'(x=\frac{\pi}{2})=40*cos(8*\frac{\pi}{2}) = 40cos(4\pi) = 40*1=40$

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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #81 : Derivative At A Point

Example Question #172 : Derivative Review

Example Question #171 : Derivative Review

Example Question #1292 : Calculus Ii

Example Question #91 : Derivative At A Point

Example Question #1301 : Calculus Ii

Example Question #1301 : Calculus Ii

Example Question #1302 : Calculus Ii

Example Question #1303 : Calculus Ii

Example Question #92 : Derivative At A Point

All Calculus 2 Resources

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