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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #1161 : Calculus Ii

Find dy/dx:

$y=\ln(\tan^{-1}(2x^2))$

Possible Answers:

$\frac{dy}{dx}=\frac{4x}{\tan^{-1}(4x)+2x^2}$

$\frac{dy}{dx}=\frac{2x}{\tan^{-1}(2x^2)}$

$\frac{dy}{dx}=\frac{2x^2}{1+4x^2}$

$\frac{dy}{dx}=\frac{2x^2}{1-4x^2\tan^{-1}(2x)}$

$\frac{dy}{dx}=\frac{4x}{(1+4x^4)\tan^{-1}(2x^2)}$

Correct answer:

$\frac{dy}{dx}=\frac{4x}{(1+4x^4)\tan^{-1}(2x^2)}$

Explanation:

Computing the derivative for this function requires us to know the following rules:

$\frac{d}{dx}[\ln(u)]=\frac{1}{u}\frac{du}{dx}$

$\frac{d}{dx}[\tan^{-1}(u)]=\frac{1}{1+u^2}\frac{du}{dx}$

We can now take the derivative of the given function:

$y=\ln(\tan^{-1}(2x^2))$

u_1=2x^2 , $\frac{du_1}{dx}=4x$

$u_2=\tan^{-1}(u_1)$ , $\frac{du_2}{dx}=\frac{1}{1+u_1^2}\frac{du_1}{dx}$

$\frac{d}{dx}[\ln(u_2)]=\frac{1}{u_2}\frac{du_2}{dx}$

$\frac{dy}{dx}=(\frac{1}{\tan^{-1}(2x^2)})(\frac{1}{1+(2x^2)^2})(4x)$

This simplifies to:

$\frac{dy}{dx}=\frac{4x}{(1+4x^2)\tan^{-1}(2x^2)}$

This is one of the answer choices.

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Example Question #31 : Derivatives

Use the definition of a derivative to find f'(x) when f(x)=3x+7 .

Possible Answers:

f'(x)=7

f'(x)=-3

f'(x)=3x

f'(x)=-3x

f'(x)=3

Correct answer:

f'(x)=3

Explanation:

In order to find f'(x) , we must remember that we can define a derivative as $f'(x)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$ . Given f(x)=3x+7 , we can set a=0 , calculate f(a)=f(0)=3(0)+7=7 and solve the limit as:

$f'(x)=\lim_{x\rightarrow 0}\frac{(3x+7)-7}{x-0}$

$f'(x)=\lim_{x\rightarrow 0}\frac{3x}{x}$

$f'(x)=\lim_{x\rightarrow 0}3$

f'(x)=3

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Example Question #34 : Definition Of Derivative

Use the definition of a derivative to find f'(x) when f(x)=7x-6 .

Possible Answers:

f'(x)=7x

f'(x)=7

f'(x)=-7x

f'(x)=-7

f'(x)=-6

Correct answer:

f'(x)=7

Explanation:

In order to find f'(x) , we must remember that we can define a derivative as $f'(x)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$ . Given f(x)=7x-6 , we can set a=0 , calculate f(a)=f(0)=7(0)-6=-6 , and solve the limit as:

$f'(x)=\lim_{x\rightarrow 0}\frac{(7x-6)-(-6)}{x-0}$

$f'(x)=\lim_{x\rightarrow 0}\frac{7x}{x}$

$f'(x)=\lim_{x\rightarrow 0}7$

f'(x)=7

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Example Question #35 : Definition Of Derivative

Use the definition of a derivative to find f'(x) when f(x)=9-5x .

Possible Answers:

f'(x)=-5x

f'(x)=9

f'(x)=5

f'(x)=-5

f'(x)=5x

Correct answer:

f'(x)=-5

Explanation:

In order to find f'(x) , we must remember that we can define a derivative as $f'(x)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$ . Given f(x)=9-5x , we can set a=0 , calculate f(a)=f(0)=9-5(0)=9 , and solve the limit as:

$f'(x)=\lim_{x\rightarrow 0}\frac{(9-5x)-(9)}{x-0}$

$f'(x)=\lim_{x\rightarrow 0}-\frac{5x}{x}$

$f'(x)=\lim_{x\rightarrow 0}-5$

f'(x)=-5

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Example Question #36 : Definition Of Derivative

Use the definition of a derivative to find f'(x) when f(x)=2-11x .

Possible Answers:

Correct answer:

Explanation:

In order to find f'(x) , we must remember that we can define a derivative as $f'(x)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$ . Given f(x)=2-11x , we can set a=0 , calculate f(a)=f(0)=2-11(0)=2 , and solve the limit as:

$f'(x)=\lim_{x\rightarrow 0}\frac{(2-11x)-2}{x-0}$

$f'(x)=\lim_{x\rightarrow 0}\frac{-11x}{x}$

$f'(x)=\lim_{x\rightarrow 0}-11$

f'(x)=-11

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Example Question #41 : Derivatives

Use the definition of a derivative to find f'(x) when $f(x)=\frac{3}{2}x+7$ .

Possible Answers:

$-\frac{3}{2}$

$\frac{3}{2}$

$\frac{2}{3}$

$-\frac{2}{3}$

Correct answer:

$\frac{3}{2}$

Explanation:

In order to find f'(x) , we must remember that we can define a derivative as $f'(x)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$ . Given $f(x)=\frac{3}{2}x+7$ , we can set a=0 , calculate $f(a)=f(0)=\frac{3}{2}(0)+7=7$ , and solve the limit as:

$f'(x)=\lim_{x\rightarrow 0}\frac{(\frac{3}{2}x+7)-7}{x-0}$

$f'(x)=\lim_{x\rightarrow 0}\frac{\frac{3}{2}x}{x}$

$f'(x)=\lim_{x\rightarrow 0}\frac{3}{2}$

$f'(x)=\frac{3}{2}$

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Example Question #42 : Derivatives

Use the definition of a derivative to find f'(x) when f(x)=-8x-5 .

Possible Answers:

Correct answer:

Explanation:

In order to find f'(x) , we must remember that we can define a derivative as $f'(x)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$ .

Given f(x)=-8x-5 , we can set a=0 , calculate f(a)=f(0)-8(0)-5=-5 , and solve the limit as:

$f'(x)=\lim_{x\rightarrow 0}\frac{(-8x-5)-(-5)}{x-0}$

$f'(x)=\lim_{x\rightarrow 0}-\frac{8x}{x}$

$f'(x)=\lim_{x\rightarrow 0}-8$

f'(x)=-8

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Example Question #43 : Derivatives

Find the velocity of a car at x=3 when the position of the car is given by the following function:

f(x)=x^2+3x+7

Possible Answers:

Correct answer:

Explanation:

The limit definition of a derivative is as follows:

$\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

where represents a very small change in .

Now, use the function given for the above formula:

$\lim_{h\rightarrow 0}\frac{(x+h)^2+3(x+h)+2-(x^2+3x+2)}{h}$

which simplified becomes

$\lim_{h\rightarrow 0}\frac{h(2x+3)}{h}=2x+3$

Finally, plug in the given point:

2(3)+3=9

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Example Question #44 : Derivatives

Find the derivative of the given function using the definition of derivative.

f(x)=x^2+x

Possible Answers:

f'(x)=3x

f'(x)=x+1

f'(x)=2x

f'(x)=2x+1

Correct answer:

f'(x)=2x+1

Explanation:

The definition of the derivative is

$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

For this problem, the derivative is

$f'(x)=\lim_{h\rightarrow 0}\frac{(x+h)^2+(x+h)-(x^2+x)}{h}$

$=\lim_{h\rightarrow 0}\frac{x^2+2xh+h^2+x+h-x^2-x}{h}$

$=\lim_{h\rightarrow 0}\frac{2xh+h^2+h}{h}$

$=\lim_{h\rightarrow 0} 2x+h+1$

As such the derivative is

f'(x)=2x+1

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Example Question #45 : Derivatives

Find dy/dx:

$y=\tan^{-1}(e^{2x^2})$

Possible Answers:

$2x^2e^{2x^2}$

$\frac{e^{2x^2}}{1+e^{4x^4}}$

$\frac{2xe^{2x^2}}{1+e^{2x^4}}$

$\frac{4xe^{2x^2}}{1+e^{4x^2}}$

$\frac{4xe^{2x^2}}{1+e^{4x^4}}$

Correct answer:

$\frac{4xe^{2x^2}}{1+e^{4x^4}}$

Explanation:

To solve for the derivative of the given function, we must realize the following:

$\frac{d}{dx}[\tan^{-1}(u)]=\frac{1}{1+u^2}\frac{du}{dx}$

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

Given:

$y=\tan^{-1}(e^{2x^2})$

$\frac{dy}{dx}=\frac{1}{1+(e^{2x^2})^2}(e^{2x^2})(4x)$

This simplifies to:

$\frac{dy}{dx}=\frac{4xe^{2x^2}}{1+e^{4x^4}}$

This is one of the answer choices.

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1161 : Calculus Ii

Find dy/dx:

Example Question #31 : Derivatives

Example Question #34 : Definition Of Derivative

Example Question #35 : Definition Of Derivative

Example Question #36 : Definition Of Derivative

Example Question #41 : Derivatives

Example Question #42 : Derivatives

Example Question #43 : Derivatives

Example Question #44 : Derivatives

Example Question #45 : Derivatives

Find dy/dx:

$y=\tan^{-1}(e^{2x^2})$

All Calculus 2 Resources

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