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Calculus 2 : Derivative Review

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Example Questions

Example Question #7 : Computation Of Derivatives

What is the rate of change of the function f(x)=1+2x+3x^4 at the point (1,6) ?

Possible Answers:

2594

Correct answer:

Explanation:

The rate of change of a function at a point is the value of the derivative at that point. First, take the derivative of f(x) using the power rule for each term.

Remember that the power rule is

$\frac{d}{dx}x^n=nx^{n-1}$ , and that the derivative of a constant is zero.

f'(x)=2+12x^3

Next, notice that the x-value of the point (1,6) is 1, so substitute 1 for x in the derivative.

f'(1)=2+12(1)^3=14

Therefore, the rate of change of f(x) at the point (1,6) is 14.

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Example Question #1 : Instantaneous Rate Of Change, Average Rate Of Change, And Linear Approximation

Calculate the derivative of f(x)=4x^2-x+2 at the point x=1 .

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.

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)=4x^2-x+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)=8x-1

Then, plug in the value of x and evaluate

f'(x=1)=8*1-1 = 7

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Example Question #2 : Instantaneous Rate Of Change, Average Rate Of Change, And Linear Approximation

Evaluate the first derivative if

f(x)=2e^x and x=0 .

Possible Answers:

f'(0)=0

f'(0)=1

f'(0)=2

f'(0)=e

Correct answer:

f'(0)=2

Explanation:

First we must find the first derivative of the function.

Because the derivative of the exponential function is the exponential function itelf, or

$\frac{\mathrm{d} }{\mathrm{d} x}[e^x]=e^x$

and taking the derivative is a linear operation,

we have that

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[f(x)]$

$=\frac{\mathrm{d} }{\mathrm{d} x}[2e^x]$

$=2\frac{\mathrm{d} }{\mathrm{d} x}[e^x]$

=2e^x

Now setting x=0

f'(0)=2e^0=2*1=2

Thus

f'(0)=2

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Example Question #41 : Limits Of Functions (Including One Sided Limits)

What is the derivative of $(2+3cos(3x))^\pi$ ?

Possible Answers:

$3\pi(2+cos(3x))^{\pi-1}sin(3x)$

$3\pi(2+cos(3x))^{\pi-1}cos(3x)$

$-3\pi(2+cos(3x))^{\pi-1}sin(3x)$

$-3\pi(2+cos(3x))^{\pi-1}$

$-3\pi(2+cos(3x))^{\pi-1}cos(3x)$

Correct answer:

$-3\pi(2+cos(3x))^{\pi-1}sin(3x)$

Explanation:

Need to use the power rule which states: $\frac{d}{dx}u^n=nu^{n-1}\frac{du}{dx}$

In our problem $\frac{du}{dx}=-3sin(3x)$

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Example Question #21 : Calculus Review

$f (x) = 7x^{2} + 6x - 9$

Give f'(x) .

Possible Answers:

f'(x) = 14x

f'(x) = 14x+12

f'(x) = 14x+6

f'(x) = 7x+6

$f'(x) = 14x^{2}+6x$

Correct answer:

f'(x) = 14x+6

Explanation:

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0, so

$f (x) = 7x^{2} + 6x - 9$

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

$f ' (x) = 14\cdot x^{1} + 6$

f ' (x) = 14x + 6

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Example Question #371 : Ap Calculus Bc

$f(x)= 8x^{3}-7x^{2}+11$

Give f''(x) .

Possible Answers:

f ''(x) = 48x -14

f ''(x) = 24x - 11

f ''(x) = 24x -14

f ''(x) = 24x

f ''(x) = 48x

Correct answer:

f ''(x) = 48x -14

Explanation:

First, find the derivative f'(x) of $f(x)= 8x^{3}-7x^{2}+11$ .

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0, so

$f(x)= 8x^{3}-7x^{2}+11$

$f'(x)= 3 \cdot 8x^{3-1}-2\cdot 7x^{2-1}+0$

$f'(x)= 24x^{2}-14x^{1}$

$f'(x)= 24x^{2}-14x$

Now, differentiate f'(x) to get f''(x) .

$f''(x)= 2 \cdot 24x^{2-1}-1\cdot 14x^{1-1}$

$f''(x)= 48x^{1}-14x^{0}$

f''(x)= 48x-14

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Example Question #23 : Calculus Review

Differentiate $x^{999}$ .

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 999x^{998}$

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 1,000x^{1,000}$

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 998x^{1,000}$

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 998x^{998}$

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 998x^{999}$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 999x^{998}$

Explanation:

$\frac{\mathrm{d} }{\mathrm{d} x} \left (x^{n} \right )= n\cdot x^{n-1}$ , so

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 999x^{999-1} = 999x^{998}$

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Example Question #1 : Computation Of Derivatives

Give the second derivative of $x^{777}$ .

Possible Answers:

$\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}} \left ( x^{777}\right ) = 602,952 x^{779}$

$\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}} \left ( x^{777}\right ) = 603,729 x^{775}$

$\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}} \left ( x^{777}\right ) = 602,952 x^{775}$

$\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}} \left ( x^{777}\right ) = 604,506 x^{779}$

$\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}} \left ( x^{777}\right ) = 603,729 x^{777}$

Correct answer:

$\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}} \left ( x^{777}\right ) = 602,952 x^{775}$

Explanation:

Find the derivative of $x^{777}$ , then find the derivative of that expression.

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , so

$\frac{\mathrm{d} }{\mathrm{d} x} \left (x^{777} \right )= 777\cdot x^{777-1} = 777x^{776}$

$\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}} \left ( x^{777}\right ) = \frac{\mathrm{d} }{\mathrm{d} x} \left ( 777x^{776} \right )= 777\cdot776\cdot x^{776-1} = 602,952x^{775}$

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Example Question #2 : Computation Of Derivatives

$f(x)= 0.8x^{3}-0.5x^{2}+ 0.9$

Give f'(x) .

Possible Answers:

$f'(x) = 2.4x^{2}-x + 0.9$

$f'(x) = 2.4x^{3}-x^{2}$

$f'(x) = 2.4x^{2}-1$

$f'(x) = 2.4x^{2}-x$

$f'(x) = 2.4x^{3}-x^{2}+ 0.9$

Correct answer:

$f'(x) = 2.4x^{2}-x$

Explanation:

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0, so

$f(x)= 0.8x^{3}-0.5x^{2}+ 0.9$

$f'(x)= 3\cdot 0.8 x^{3-1}-2\cdot 0.5x^{2-1}+ 0$

$f'(x)= 2.4x^{2}-1x^{1}$

$f'(x) = 2.4x^{2}-x$

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Example Question #24 : Calculus Review

$f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$

Give f''(x) .

Possible Answers:

$f''(x) =10.8x^{2} + 8.4x - 0.5$

$f''(x) =10.8x^{2} + 8.4x$

$f''(x) =10.8x^{2} + 4.2x$

$f''(x) =10.8x^{2} + 2.1x$

$f''(x) =10.8x^{2} + 4.2x - 0.5$

Correct answer:

$f''(x) =10.8x^{2} + 4.2x$

Explanation:

First, find the derivative f'(x) of $f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$ .

Recall that $\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0.

$f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$

$f'(x) =4\cdot 0.9x^{4-1} + 3\cdot 0.7x^{3-1}- 1 \cdot 0.5 x ^{1-1}$

$f'(x) =3.6x^{3} + 2.1x^{2}- 0.5 x ^{0}$

$f'(x) =3.6x^{3} + 2.1x^{2}- 0.5$

Now, differentiate f'(x) to get f''(x) .

$f'(x) =3.6x^{3} + 2.1x^{2}- 0.5$

$f''(x) =3 \cdot 3.6x^{3-1} + 2 \cdot 2.1x^{2-1}- 0$

$f''(x) =10.8x^{2} + 4.2x^{1}- 0$

$f''(x) =10.8x^{2} + 4.2x$

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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #7 : Computation Of Derivatives

Example Question #1 : Instantaneous Rate Of Change, Average Rate Of Change, And Linear Approximation

Example Question #2 : Instantaneous Rate Of Change, Average Rate Of Change, And Linear Approximation

Example Question #41 : Limits Of Functions (Including One Sided Limits)

Example Question #21 : Calculus Review

Example Question #371 : Ap Calculus Bc

Example Question #23 : Calculus Review

Example Question #1 : Computation Of Derivatives

Example Question #2 : Computation Of Derivatives

Example Question #24 : Calculus Review

All Calculus 2 Resources

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