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Calculus 2 : Derivative Review

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Example Questions

Example Question #41 : Derivative Review

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

Possible Answers:

$\frac{3}{2}$

$-\frac{2}{3}$

$\frac{2}{3}$

$-\frac{3}{2}$

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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Example Question #46 : Derivatives

Find dy/dx:

$y=\ln(e^{4x^2-\tan^{-1}(2x)})$

Possible Answers:

$\frac{32x^3+8x-2}{1+4x^2}$

$\frac{8x-\frac{2}{1+2x^2}}{4x^2-\tan^{-1}(2x)}$

$\frac{12x^3+8x+2}{1+4x^2}$

$e^{4x^2-\tan^{-1}(2x)}$

$\frac{8x-2}{4x^2-\tan^{-1}(2x)}$

Correct answer:

$\frac{32x^3+8x-2}{1+4x^2}$

Explanation:

Before attempting to take the derivative of the given function, the following identity must be realized:

$\ln e^a=a\cdot\ln e=a(1)=a$

Using this identity, the given function can be simplified to:

$y=4x^2-\tan^{-1}(2x)$

From this taking the derivative is a fairly straightforward process:

$\frac{dy}{dx}=8x-\frac{1}{1+(2x^2)}(2)=8x-\frac{2}{1+4x^2}$

We can simplify this expression by placing all the terms under a common denominator like so:

$8x(\frac{1+4x^2}{1+4x^2})-\frac{2}{1+4x^2}=\frac{8x+32x^3-2}{1+4x^2}=\frac{32x^3+8x-2}{1+4x^2}$

This is one of the answer choices.

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Example Question #47 : Derivatives

Use a definition of the derivative with the function f(x) = a^x to evaluate the following limit:

$\lim_{h \to 0}\frac{a^h-1}{h}$

Possible Answers:

1/a

$\ln(a)$

Correct answer:

$\ln(a)$

Explanation:

Using the definition $f'(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$

And plugging in our function, we get that

$(a^x)' = \lim_{h \to 0}\frac{a^{x+h}-a^x}{h}$ .

if we factor out a^x inside the limit we get $\lim_{h\to 0} a^x(\frac{a^h-1}{h})$

since the a^x term doesn't contain an h we can factor it out, and then divide by both sides, getting that

$\frac{(a^x)'}{a^x} = \lim_{h\to 0}\frac{a^h-1}{h}$

but we know that $(a^x)' = a^x \cdot \ln(a)$

So we find that the limit is equal to $\ln(a)$ .

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Example Question #48 : Derivatives

If the definition of the derivative is used to find the derivative of $x\sin(x)$ . Which of these expressions must be evaluated?

Possible Answers:

None of the other answers

$\lim_{h \to 0} \frac{(x+h)\sin(x)-x\sin(x+h)}{h}$

$\lim_{h \to 0} \frac{x\sin(x)+x\sin(h)}{h}$

$\lim_{h \to 0} \frac{(x+h)\sin(x+h)-x\sin(x)}{h}$

$\lim_{h \to 0} \frac{(x+h)\sin(x+h)-x\sin(x)}{xh}$

Correct answer:

$\lim_{h \to 0} \frac{(x+h)\sin(x+h)-x\sin(x)}{h}$

Explanation:

The definition of the derivative of a function is given by

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

With $f(x) = x\sin(x)$ , we substitute x+h into our function, and get $f(x+h) = (x+h)\sin(x+h)$ . Substituting these into the formula gives us

$\lim_{h \to 0} \frac{(x+h)\sin(x+h)-x\sin(x)}{h}$ .

Note: this limit is not easy to evaluate by hand, it would be easier to find the derivative using the Product Rule.

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Example Question #49 : Derivatives

Possible Answers:

f'(x)=x^2cos(x)+2xsin(x)

f''(x)=x^2sin(x)+4xcos(x)+2sin(x)

f'(x)=x^2cos(x)+2xsin(x)

f''(x)=-x^2sin(x)+4xcos(x)+2sin(x)

f'(x)=x^2sin(x)+2xcos(x)

f''(x)=-x^2cos(x)+4xsin(x)+2cos(x)

f'(x)=2xcos(x)

f''(x)=2sin(x)

f'(x)=2xcos(x)

f''(x)=-2sin(x)

Correct answer:

f'(x)=x^2cos(x)+2xsin(x)

f''(x)=-x^2sin(x)+4xcos(x)+2sin(x)

Explanation:

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Example Question #1171 : Calculus Ii

Find the derivative of $f(x)=5x^{2}+12x-7$ using the definition of the derivative.

Possible Answers:

f'(x)=10

f'(x)=5x+12

f'(x)=10x+12

f'(x)=5x^2+12x-7

Correct answer:

f'(x)=10x+12

Explanation:

The definition of a derivative is $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f(x)=5x^{2}+12x-7$

$f(x+h)=5(x+h)^{2}+12(x+h)-7$

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{5(x+h)^{2}+12(x+h)-7-(5x^{2}+12x-7)}{h}$

$=\lim_{h\rightarrow 0}\frac{5(x^{2}+2xh+h^{2})+12x+12h-7-5x^{2}-12x+7}{h}$

$=\lim_{h\rightarrow 0}\frac{5x^{2}+10xh+5h^{2}+12x+12h-7-5x^{2}-12x+7}{h}$

$=\lim_{h\rightarrow 0}\frac{10xh+5h^{2}+12h}{h}$

$=\lim_{h\rightarrow 0}\frac{h(10x+5h+12)}{h}$

$=\lim_{h\rightarrow 0}10x+5h+12$

=10x+5(0)+12

f'(x)=10x+12

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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #41 : Derivative Review

Example Question #42 : Derivatives

Example Question #43 : Derivatives

Example Question #44 : Derivatives

Example Question #45 : Derivatives

Find dy/dx:

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

Example Question #46 : Derivatives

Find dy/dx:

Example Question #47 : Derivatives

Example Question #48 : Derivatives

Example Question #49 : Derivatives

Example Question #1171 : Calculus Ii

All Calculus 2 Resources

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