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Calculus 3 : Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Velocity, Speed, Acceleration

Let

$f(x)=e^{\pi x}$

Find the first and second derivative of the function.

Possible Answers:

$f'(x)=e^{\pi }$

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

$f'(x)=\pi$

$f''(x)=\0$

$f'(x)=\pi e^{\pi x }$

$f''(x)=\pi^2 e^{\pi x}$

$f'(x)=e^{\pi x }$

$f''(x)=e^{\pi x}$

Correct answer:

$f'(x)=\pi e^{\pi x }$

$f''(x)=\pi^2 e^{\pi x}$

Explanation:

In order to solve for the first and second derivative, we must use the chain rule.

The chain rule states that if

y=f(u)

and

u=g(x)

then the derivative is

$\frac{dy}{dx}=\frac{dy}{du}*\frac{du}{dx}$

In order to find the first derviative of the function

$f(x)=e^{\pi x}$

we set

y=e^u

and

$u=\pi x$

Because the derivative of the exponential function is the exponential function itself, we get

$\frac{dy}{du}=\frac{d}{du}[e^u]$

$\frac{dy}{du}=e^u$

And differentiating we use the power rule which states

$\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

As such

$\frac{du}{dx}=\frac{\mathrm{d} }{\mathrm{d} x}[\pi x]$

$\frac{du}{dx}=\pi x^{1-1}=\pi$

And so

$\frac{dy}{dx}=e^{\pi x}*\pi=\pi e^{\pi x}$

$f'(x)=\pi e^{\pi x}$

To solve for the second derivative we set

$y=\pi e^{u}$

and

$u=\pi x$

Because the derivative of the exponential function is the exponential function itself, we get

$\frac{dy}{du}=\frac{d}{du}[\pi e^u]$

$\frac{dy}{du}=\pi e^u$

And differentiating we use the power rule which states

$\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

As such

$\frac{du}{dx}=\frac{\mathrm{d} }{\mathrm{d} x}[\pi x]$

$\frac{du}{dx}=\pi x^{1-1}=\pi$

And so the second derivative becomes

$\frac{dy}{dx}=\pi e^{\pi x}*\pi=\pi^2 e^{\pi x}$

$f''(x)=\pi^2e^{\pi x}$

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Example Question #31 : Derivatives

Find the second derivative of the following function:

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

Possible Answers:

$\cos(x)(6x-x^3)-3x^2\sin(x)+4e^{2x}\:$

$\cos(x)(6x-x^3)-6x^2\sin(x)+2e^{2x}$

$\cos(x)(6x-x^3)+6x^2\sin(x)+4e^{2x}$

$\cos(x)(6x-x^3)-6x^2\sin(x)+4e^{2x}$

$\cos(x)(6x+x^3)-6x^2\sin(x)+4e^{2x}$

Correct answer:

$\cos(x)(6x-x^3)-6x^2\sin(x)+4e^{2x}$

Explanation:

The first derivative of the function is equal to

$f'(x)=3x^2\cos(x)-x^3\sin(x)+2e^{2x}$

The second derivative - the derivative of the function above - of the original function is equal to

$f''(x)=\cos(x)(6x-x^3)-6x^2\sin(x)+4e^{2x}$

Both derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{du} }{\mathrm{d} x}$

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Example Question #221 : Calculus 3

Calculate the derivative of the function given below

$f(x) = \left(x^{3}-\sin {x} \right ) ^ 4$

Possible Answers:

$4 \left( x^3 - \sin {x} \right ) ^3 \left( 3x ^ 2 + \sin x \right )$

$4 \left( x^3 - \sin {x} \right ) ^3 \left( 3x ^ 2 - \cos x \right )$

$4 \left( 3x ^ 2 - \cos x \right ) ^ 3$

$4 \left( x^3 - \sin {x} \right ) ^3 \left( 3x - \cos x \right )$

Correct answer:

$4 \left( x^3 - \sin {x} \right ) ^3 \left( 3x ^ 2 - \cos x \right )$

Explanation:

We have a function inside of another function, therefore we have a chain-rule.

$f' (x) =4 \left( x^3 - \sin {x} \right ) ^3 \left( 3x ^ 2 - \cos x \right )$ .

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Example Question #222 : Calculus 3

In order to find the maximas and minimas of a given function f (x) , one must set which quantity equal to zero $\left(0 \right )$ and solve for ?

Possible Answers:

f (x) = 0

f'' (x) = 0

f' (x) = 0

-f (x) = 0

Correct answer:

f' (x) = 0

Explanation:

At the location of a maximum or minimum of a function, the tangent line has a slope of zero $\left(0 \right )$ . We know the first derivative, f' (x) , of a function f (x) , gives us the value of the tangent line for a given value of . Therefore, one needs to set the first derivative f' (x) = 0 .

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Example Question #223 : Calculus 3

Find the slope of the line tangent to f(x) , given below, when x=4 .

$f(x)= 2 \sqrt{x^2+4x-2}$

Possible Answers:

f'(2)=4

$f'(2)=2\sqrt{2}$

$f'(2)=2\sqrt{8}$

$f'(2)=4\sqrt{2}$

Correct answer:

$f'(2)=2\sqrt{2}$

Explanation:

First, we need to find the derivative of f(x) , given below as

$f'(x)= \frac{2x+4}{\sqrt{x^2+4x-4}}$ . Plugging a value of x=2 gives us

$f'(2)= \frac{2(2)+4}{\sqrt{2^2+4(2)-4}}=\frac{8}{\sqrt{8}}=\frac{8}{\sqrt{2\cdot4}}=\frac{4}{\sqrt{2}}=2\sqrt{2}$ .

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Example Question #222 : Calculus 3

Find the derivative of the function $y= e^{2a}$ , where is any constant.

Possible Answers:

$2e^{2a}$

$2ae^{2a}$

ae^a

None of the other answers

2ae^a

Correct answer:

None of the other answers

Explanation:

The correct answer is

Notice that this function is constant; it does not contain any variables. ( is stated to be a constant). Hence the derivative is .

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Example Question #33 : Derivatives

Find the derivative of the function $f(x) = 6x^a+e^{\sin(x)} +a\sin(ax)$ , where is any constant.

Possible Answers:

$\cos(x)e^{\sin(x)}$

$6+\cos(x)e^{\sin(x)}$

$6+\cos(x)e^{\sin(x)}+a\cos(ax)$

None of the other answers

$6ax^{a-1}+\cos(x)e^{\sin(x)}+a^2\cos(ax)$

Correct answer:

$6ax^{a-1}+\cos(x)e^{\sin(x)}+a^2\cos(ax)$

Explanation:

Looking at the term 6x^a , since is a constant, the derivative is $a\times 6x^{a-1} = 6ax^{a-1}$ .

The derivative of $e^{\sin(x)}$ uses the Chain Rule. So the derivative is $e^{\sin(x)} \times (\sin(x))' = \cos(x)e^\sin(x)$ .

Remembering that is just constant, we have the derivative of $a\sin(ax)$ is $a\cos(ax) \times a = a^2\cos(ax)$ , also by the chain rule.

Adding our three results together gives

$6ax^{a-1}+\cos(x)e^{\sin(x)}+a^2\cos(ax)$ .

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Example Question #34 : Calculus Review

Evaluate the derivative of the function $f(t) = a\sin(t)$ , where is any constant.

Possible Answers:

None of the other answers

$\sin(t)$

$a\cos(t)$

$a\cos(t)+\sin(t)$

$\cos(t)$

Correct answer:

$a\cos(t)$

Explanation:

It's important to take note that is not a variable, it is a constant. Hence the derivative of is , not . This idea will pop up in many places later on in Calculus 3.

Either the Product Rule, or the Constant Multiple Rule can be used to find the derivative of $a\sin(t)$ .

For example, the Product Rule will proceed as follows

$(a\sin(t))' = (a)(\sin(t))'+(a)'(\sin(t))$

$= (a)(\cos(t))+(0)(\sin(t)) = a\cos(t)$ .

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Example Question #227 : Calculus 3

Find $f' \left( x \right )$ of $f \left(x \right )= \tan \left(3x^2-5x \right )$ .

Possible Answers:

$\sec ^2 \left(3x^2-5x \right ) \cdot \left(6x-5 \right ) \cdot 6$

$\sec \left(3x^2-5x \right ) \cdot \left(6x-5 \right )$

$\sec ^2 \left(3x^2-5x \right )$

$\sec ^2 \left(6x-5 \right )$

$\sec ^2 \left(3x^2-5x \right ) \cdot \left(6x-5 \right )$

Correct answer:

$\sec ^2 \left(3x^2-5x \right ) \cdot \left(6x-5 \right )$

Explanation:

Note that $\frac{d}{dx} \tan \left( f(x) \right )= \sec ^2 \left(f(x) \right ) \cdot f'(x)$ .

For this problem, $f(x) = 3x^2 - 5x\Rightarrow f'(x)=6x - 5$ . Putting this together with the definition, we arrive at

$f' (x)= \sec ^2 \left(3x^2-5x \right ) \cdot \left(6x-5 \right )$ .

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Example Question #228 : Calculus 3

Calculate f' (1) of $f(x)= \ln \left(\frac{6x^2}{\left(x+4 \right )} \right )$ .

Possible Answers:

$-\frac{2}{15}$

$\frac{9}{5}$

$\ln 6$

Correct answer:

$\frac{9}{5}$

Explanation:

We begin by rewriting our function in a more convenient way,

$f(x) = \ln \left(\frac{6x^2}{\left(x+4 \right )} \right )= \ln \left(6x^2 \right )- \ln \left(x+4 \right )$ .

This makes taking the derivative a little easier. We can therefore write

$f' (x) = \frac{2}{x}-\frac{1}{x+4}$ , or $f' (1) = 2 - \frac{1}{5} = \frac{9}{5}$ .

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1 : Velocity, Speed, Acceleration

Example Question #31 : Derivatives

Example Question #221 : Calculus 3

Example Question #222 : Calculus 3

Example Question #223 : Calculus 3

Example Question #222 : Calculus 3

Example Question #33 : Derivatives

Example Question #34 : Calculus Review

Example Question #227 : Calculus 3

Example Question #228 : Calculus 3

All Calculus 3 Resources

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