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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 #1571 : Calculus Ii

What is the derivative of f(x)=4x^3-9x^2+7x-1 ?

Possible Answers:

$f{}'(x)=12x^2-18x-7$

$f{}'(x)=12x^2-8x+7$

$f{}'(x)=12x^2-18x+7$

$f{}'(x)=12x^2+18x+7$

$f{}'(x)=x^2-18x+7$

Correct answer:

$f{}'(x)=12x^2-18x+7$

Explanation:

Recall that when taking the derivative, multiply the exponent by the coefficient in front of the x term and then also subtract one from the exponent:

f(x)=4x^3-9x^2+7x-1

$f'(x)=3\cdot 4x^{3-1}-2\cdot9x^{2-1}+1\cdot7x^{1-1}$

$f{}'(x)=12x^2-18x+7$

Recall that the derivative of a constant is just zero.

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Example Question #1572 : Calculus Ii

What is the second derivative of f(x)=16x^2-5x+9 ?

Possible Answers:

f{}''(x)=3

f{}''(x)=32x-5

f{}''(x)=32

f{}''(x)=32x

f{}''(x)=2x

Correct answer:

f{}''(x)=32

Explanation:

Before you can take the second derivative, you must first take the first derivative. Remember to multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent:

f{}'(x)=32x-5

Now, take the second derivative:

f{}''(x)=32

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Example Question #251 : First And Second Derivatives Of Functions

Find the first and second derivative of the function.

$f(x)=e^{5x}$

Possible Answers:

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

$f''(x)=25e^{5x}$

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

$f'(x)=5xe^{5x}$

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

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

$f'(x)=5xe^{5x}$

$f''(x)=25x^2e^{5x}$

Correct answer:

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

$f''(x)=25e^{5x}$

Explanation:

To find the first and second derivatives, we use the chain rule.

F'(x)=f'(g(x))g'(x)

where F(x)=f(g(x)) .

For the first derivative we have f(x)=e^x and g(x)=5x and so

$f'(x)=e^{5x}*5=5e^{5x}$

For the second derivative we have f(x)=5e^x and g(x)=5x and so

$f''(x)=5e^{5x}*5=25e^x$

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Example Question #251 : First And Second Derivatives Of Functions

Find the first and second derivative of the function.

$f(x)=x+sin\,x$

Possible Answers:

f'(x)=0

$f''(x)=-sin\,x$

$f'(x)=1+cos\,x$

$f''(x)=-sin\,x$

$f'(x)=cos\,x$

$f''(x)=sin\,x$

f'(x)=1

f''(x)=0

Correct answer:

$f'(x)=1+cos\,x$

$f''(x)=-sin\,x$

Explanation:

We find the first derivative by deriving term by term. As such,

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[x]+\frac{\mathrm{d} }{\mathrm{d} x}[sin\,x]=1+cos\,x$

Taking the second derivative

$f''(x)=\frac{\mathrm{d} }{\mathrm{d} x}[1]+\frac{\mathrm{d} }{\mathrm{d} x}[cos\,x]=0-sin\,x=-sin\,x$

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Example Question #1575 : Calculus Ii

What is the first derivative of f(x)=4x^3-2x^2+x-1?

Possible Answers:

f{}'(x)=12x+1

f{}'(x)=24x-4

$f{}'(x)=12x^2-4x-1$

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

$f{}'(x)=12x^2-4x$

Correct answer:

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

Explanation:

To find the first derivative, look at each term separately.

For the first term, multiply the exponent by the coefficient ( $4\cdot3$ ) and then subtract one from the exponent to get 12x^2 .

Do the same with the next term: $2\cdot-2$ and then subtract one from the exponent to get -4x .

For the third term, the derivative is just the coefficient (1 in this case).

The derivative of a constant is 0.

Put those all together to get your answer: $f{}'(x)=12x^2-4x+1$ .

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Example Question #1581 : Calculus Ii

Find the second derivative of f(x)=2x^4-3x^3+5x^2-1 .

Possible Answers:

$f{}''(x)=4x^2-18x+10$

$f{}''(x)=24x^2-18x$

$f{}''(x)=24x^2-18x+10$

$f{}''(x)=24x^2-18x-1$

$f{}''(x)=24x^2+10$

Correct answer:

$f{}''(x)=24x^2-18x+10$

Explanation:

Before finding the second derivative, you must find the first derivative. Remember to multiply the exponent by the coefficient and then subtract one from the exponent. Evaluate each term separately and then put all together at the end.

The first derivative is: $f{}'(x)=8x^3=9x^2+10x$ .

Now, take the derivative of the 1st derivative to find the second derivative:

$f{}''(x)=24x^2-18x+10$ .

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Example Question #1582 : Calculus Ii

Find the first derivative of $f(x)=\frac{6x^3-8x^2}{x^2}$ .

Possible Answers:

f{}'(x)=6

$f{}'(x)=\frac{6}{x}$

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

f{}'(x)=6x

f{}'(x)=6x+1

Correct answer:

f{}'(x)=6

Explanation:

First, chop up the fraction into 2 separate terms and simplify:

$\frac{6x^3-8x^2}{x^2}=\frac{6x^3}{x^2}-\frac{8x^2}{x^2}=6x-8$

Now, take the derivative of that expression. Remember to multiply the exponent by the coefficient and then decrease the exponent by 1:

f{}'(x)=6

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Example Question #251 : First And Second Derivatives Of Functions

$\int \frac{5}{x}-1dx$

Possible Answers:

$5ln\left | x\right |-x$

$5ln\left | x\right |-2x+C$

$5ln\left | x\right |-5x+C$

$5ln\left | x\right |-x+C$

$ln\left | x\right |-x+C$

Correct answer:

$5ln\left | x\right |-x+C$

Explanation:

Integrate each term separately. When there is just an x on the denominator, the integral of that is $ln\left | x\right |$ .

Therefore, the first term integrated is:

$5ln\left | x\right |$

The second term integrated is:

Put those together to get:

$5ln\left | x\right |-x$

Because it's an indefinite integral, remember to add C at the end:

$5ln\left | x\right |-x+C$ .

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Example Question #1 : Applications Of Derivatives

Given the velocity function

$v(x)=\int_{0}^{x^2+2x+1} z^d\sin(z^5) dz$

where is real number such that $d\in [0,1]$ , find the acceleration function

a(x) .

Possible Answers:

$a(x)=(x+1)^{d}\sin (x+1)^{10}$

$a(x)=(2x+2)(x+1)^{2d}\sin (x+1)^{10}$

$a(x)=(x^2+2x+1)^{d}\sin (x^2+2x+1)^{10}$

$a(x)=(x^2+2x+1)^{2d}\sin (x^2+2x+1)^{10}$

$a(x)=(2x+2)^{d}(x+1)^{2d}\sin (x+1)^{10}$

Correct answer:

$a(x)=(2x+2)(x+1)^{2d}\sin (x+1)^{10}$

Explanation:

We can find the acceleration function a(x) from the velocity function by taking the derivative:

a(x)=v'(x)

We can view the function

$v(x)=\int_{0}^{x^2+2x+1} z^d\sin(z^5) dz$

as the composition of the following functions

$g(x)=\int_{0}^{x} z^d\sin(z^5) dz$

h(x)=x^2+2x+1=(x+1)^2

so that v(x)=g(h(x)) . This means we use the chain rule

[g(h(x))]'=g'(h(x))h'(x)

to find the derivative. We have $g'(x)=x^d\sin x^5$ and h'(x)=2x+2 , so we have

$a(x)=v'(x)=[(x+1)^2]^d\sin [(x+1)^2]^5\cdot (2x+2)$

$=(2x+2)(x+1)^{2d}\sin (x+1)^{10}$

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Example Question #451 : Derivative Review

Define $f (x,y) = \frac{1}{x^{2}y^{2}}$ .

Find $f _{xy} (x,y)$ .

Possible Answers:

$f _{xy} (x,y) = \frac{9}{x^{3}y^{3} }$

$f _{xy} (x,y) = - \frac{9}{x^{3}y^{3} }$

$f _{xy} (x,y) = \frac{4}{x^{3}y^{3} }$

$f _{xy} (x,y) = \frac{1}{x^{3}y^{3} }$

$f _{xy} (x,y) = - \frac{4}{x^{3}y^{3} }$

Correct answer:

$f _{xy} (x,y) = \frac{4}{x^{3}y^{3} }$

Explanation:

First, find $f_{x} = \frac{\partial f}{\partial x}$ as follows:

$f (x,y) = \frac{1}{x^{2}y^{2}}$

$\frac{\partial }{\partial x}f (x,y) = \frac{\partial }{\partial x}\left ( \frac{1}{x^{2}y^{2}} \right )$

$=\frac{1}{ y^{2}} \cdot \frac{\partial }{\partial x}\left ( \frac{1}{x^{2} } \right )$

$=\frac{1}{ y^{2}} \cdot \frac{\partial }{\partial x}x^{-2}$

$=\frac{1}{ y^{2}} \cdot (-2) x^{-3} = -\frac{2}{x^{3}y^{2}}$

Therefore, $f_{x} (x,y ) = -\frac{2}{x^{3}y^{2}}$

Now, find $f_{xy} = \frac{\partial }{\partial y} f_{x}$ as follows:

$f_{x} (x,y ) = -\frac{2}{x^{3}y^{2}}$

$\frac{\partial }{\partial y} f_{x} (x,y ) = \frac{\partial }{\partial y}\left ( -\frac{2}{x^{3}y^{2}} \right )$

$= -\frac{2}{x^{3} } \cdot \frac{\partial }{\partial y}\left ( \frac{1}{ y^{2}} \right )$

$= -\frac{2}{x^{3} } \cdot \frac{\partial }{\partial y} y^{-2}$

$= -\frac{2}{x^{3} } \cdot (-2) y^{-3}$

$= -\frac{2}{x^{3} } \cdot \left ( -\frac{2}{y^{3} } \right )$

$= \frac{4}{x^{3}y^{3} }$

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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1571 : Calculus Ii

Example Question #1572 : Calculus Ii

Example Question #251 : First And Second Derivatives Of Functions

Example Question #251 : First And Second Derivatives Of Functions

Example Question #1575 : Calculus Ii

Example Question #1581 : Calculus Ii

Example Question #1582 : Calculus Ii

Example Question #251 : First And Second Derivatives Of Functions

Example Question #1 : Applications Of Derivatives

Example Question #451 : Derivative Review

All Calculus 2 Resources

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