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Calculus 2 : Derivative Review

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Example Questions

Example Question #1 : Derivatives

Evaluate the limit using one of the definitions of a derivative.

$\lim_{x \rightarrow -\frac{\pi}{2}} \frac{\sin^2x -1}{x+\frac{\pi}{2}}$

Possible Answers:

Does not exist

$\frac{1}{2}$

$\pi$

Correct answer:

Explanation:

Evaluating the limit directly will produce an indeterminant solution of $\small \tiny{\frac{0}{0}}$ .

The limit definition of a derivative is $\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=f'(x)$ . However, the alternative form, $\lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}=f'(x)$ , better suits the given limit.

Let $f(x)=\sin^2x$ and notice $f\left(-\frac{\pi}{2}\right)=1$ . It follows that ${\lim_{x \rightarrow -\frac{\pi}{2}} \frac{\sin^2-1}{x-\left(-\frac{\pi}{2} \right )} = \frac{d}{dx}\sin^2x}$ .

Thus, the limit is ${2\sin\left(-\frac{\pi}{2} \right ) \cos\left(-\frac{\pi}{2}\right)} = 0.$

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Example Question #2 : Derivatives

Evaluate the limit using one of the definitions of a derivative.

$\lim_{x \rightarrow 1} \frac{x^{49}-1}{x-1}$

Possible Answers:

Does not exist

Correct answer:

Explanation:

Evaluating the derivative directly will produce an indeterminant solution of $\small \tiny{\frac{0}{0}}$ .

Let $f(x)=x^{49}$ and notice f(1)=1 . It follows that ${\lim_{x \rightarrow 1} \frac{x^{49}-1}{x-1} = \frac{d}{dx}x^{49}}$ . Thus, the limit is $49(1)^{48}=49$ .

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Example Question #3 : Derivatives

Suppose and are differentiable functions, and f(3)=1, f'(3)=4, g(-1)=3, g'(-1)=-1 . Calculate the derivative of h(x) = f(x+2)g(-x) , at x = 1

Possible Answers:

None of the other answers

Correct answer:

None of the other answers

Explanation:

The correct answer is 11.

Taking the derivative of h(x) involves the product rule, and the chain rule.

h'(x)=[f(x+2)][-g'(-x)]+[g(-x)][f'(x+2)]

= -g'(-x)f(x+2)+g(-x)f'(x+2)

Substituting x=1 into both sides of the derivative we get

$h'(1)= -g'(-1)f(3)+g(-1)f'(3) = -1\cdot1 + 3 \cdot 4 = 11$ .

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Example Question #4 : Definition Of Derivative

Evaluate the limit

$\small \small \lim_{\mu\to 0} \frac{\tan \mu}{\mu}$

without using L'Hopital's rule.

Possible Answers:

$\small \frac{\sqrt{2}}{2}$

$\small -\frac{\sqrt{2}}{2}$

$\small 2$

$\small 1$

Correct answer:

$\small 1$

Explanation:

If we recall the definition of a derivative of a function $\small f(x)$ at a point $\small x$ , one of the definitions is

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

If we compare this definition to the limit

$\small \small \lim_{\mu\to 0} \frac{\tan \mu}{\mu}$

we see that that this is the limit definition of a derivative, so we need to find the function $\small f(x)$ and the point $\small x$ at which we are evaluating the derivative at. It is easy to see that the function is $\small f(x)=\tan x$ and the point is $\small x=0$ . So finding the limit above is equivalent to finding $\small f'(0)$ .

We know that the derivative is $\small f'(x)=\sec^2(x)$ , so we have

$\small \small \small \lim_{\mu\to 0} \frac{\tan \mu}{\mu}=f'(0)=\sec^2(0)=1^2=1$ .

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Example Question #4 : Definition Of Derivative

Approximate the derivative if $y=\sqrt{x}$ where x=2 .

Possible Answers:

0.35359

0.35355

0.35351

0.35346

0.35353

Correct answer:

0.35355

Explanation:

Write the definition of the limit.

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

Substitute x=2 .

$\lim_{h\to 0}\frac{f(2+h)-f(2)}{h}=\lim_{h\to 0}\frac{\sqrt{2+h}-\sqrt2}{h}$

Since is approaching to zero, it would be best to evaluate when we assume that is progressively decreasing. Let's assume h=0.1, 0.0001, 0.000001 and check the pattern.

$\frac{\sqrt{2+0.1}-\sqrt2}{0.1}=0.349241$

$\frac{\sqrt{2+0.0001}-\sqrt2}{0.0001}=0.353549$

$\frac{\sqrt{2+0.000001}-\sqrt2}{0.000001}=0.353553$

The best answer is: 0.35355

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Example Question #5 : Definition Of Derivative

Given:

f(x)=3x^2(x^2-5x+2)^5

Find f'(x):

Possible Answers:

(15x^2)(x^2-5x+2)^4

(x^2-5x+2)^4(36x^3-45x^2+12x)

(x^2-5x+2)^4(60x^2-150)

(x^2-5x+2)^5(36x^3-105x^2+12x)

(x^2-5x+2)^4(36x^3-105x^2+12x)

Correct answer:

(x^2-5x+2)^4(36x^3-105x^2+12x)

Explanation:

Computation of the derivative requires the use of the Product Rule and Chain Rule.

The Product Rule is used in a scenario when one has two differentiable functions multiplied by each other:

f(x)*g(x)

$\frac{d}{dx}[f(x)*g(x)]=f(x)*g'(x)+g(x)*f(x)$

This can be easily stated in words as: "First times the derivative of the second, plus the second times the derivative of the first."

In the problem statement, we are given:

f(x)=3x^2(x^2-5x+2)^5

3x^2 is the "First" function, and (x^2-5x+2)^5 is the "Second" function.

The "Second" function requires use of the Chain Rule.

When:

$y=(f(x))^a \rightarrow \frac{dy}{dx}=a*(f(x))^{a-1}*f'(x)$

Applying these formulas results in:

f'(x)=(3x^2)[5(x^2-5x+2)^4(2x-5)]+(x^2-5x+2)^5(6x)

Simplifying the terms inside the brackets results in:

f'(x)=(3x^2)[(10x-25)(x^2-5x+2)^4]+(x^2-5x+2)^5(6x)

We notice that there is a common term that can be factored out in the sets of equations on either side of the "+" sign. Let's factor these out, and make the equation look "cleaner".

f'(x)=(x^2-5x+2)^4[(3x^2)(10x-25)+(6x)(x^2-5x+2)]

Inside the brackets, it is possible to clean up the terms into one expanded function. Let us do this:

f'(x)=(x^2-5x+2)^4[30x^3-75x^2+6x^3-30x^2+12x]

Simplifying this results in one of the answer choices:

f'(x)=(x^2-5x+2)^4(36x^3-105x^2+12x)

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Example Question #4 : Definition Of Derivative

What is the value of the limit below?

$\lim_{x\to \frac{\pi}{6}}\frac{sin(x)-sin(\frac{\pi}{6})}{x-\frac{\pi}{6}}$

Possible Answers:

$\frac{\sqrt3}{2}$

$\frac{1}{2}$

Correct answer:

$\frac{\sqrt3}{2}$

Explanation:

Recall that one definition for the derivative of a function $f(x)\ at\ x=a$ is $f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$ .

This means that this question is asking us to find the value of the derivative of sin(x) at $x=\frac{\pi}{6}$ .

Since

$\frac{d}{dx}sin(x)=cos(x)$ and $cos(\frac{\pi}{6})=\frac{\sqrt3}{2}$ , the value of the limit is $\frac{\sqrt3}{2}$ .

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Example Question #3 : Derivatives

$f(x)=x^2\tan^{-1}(3x^5)$

$\frac{dy}{dx}=?$

Possible Answers:

$2x\tan^{-1}(3x^5)$

$\frac{15x^4\tan^{-1}(3x^5)}{1+9x^{10}}+2x^3$

$\frac{15x^6}{1+9x^{10}}+2x\tan^{-1}(3x^5)$

$\frac{15x^8}{1+9x^7}+2x\tan^{-1}(3x^5)$

$\frac{15x^4\tan^{-1}(3x^5)}{1+9x^{7}}+2x^3$

Correct answer:

$\frac{15x^6}{1+9x^{10}}+2x\tan^{-1}(3x^5)$

Explanation:

Evaluation of this integral requires use of the Product Rule. One must also need to recall the form of the derivative of $\tan^{-1}(x)$ .

Product Rule:

$\frac{d}{dx}[f(x)g(x)]=f(x)g'(x)+g(x)f'(x)$

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

Applying these two rules results in:

$\frac{dy}{dx}=\frac{d}{dx}[x^2\tan^{-1}(3x^5)]$

$\frac{dy}{dx}=(x^2)(\frac{1}{1+(3x^5)^2})(15x^4)+\tan^{-1}(3x^5)(2x)$

$\frac{dy}{dx}=\frac{15x^6}{1+9x^{10}}+2x\tan^{-1}(3x^5)$

This matches one of the answer choices.

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Example Question #3 : Derivative Review

Use the definition of the derivative to solve for f'(x) .

f(x)=x^3

Possible Answers:

$\frac{1}{3}x^2$

3x^2

x^2

Correct answer:

3x^2

Explanation:

In order to find f'(x) , we need to remember how to find f'(x) by using the definition of derivative.

Definition of Derivative:

$f'(x)=\lim_{h\rightarrow0} \frac{f(x+h)-f(x)}{h}$

Now lets apply this to our problem.

$f'(x)=\lim_{h\rightarrow0}\frac{(x+h)^3-x^3}{h}$

Now lets expand the numerator.

$f'(x)=\lim_{h\rightarrow0}\frac{x^3+3x^2h+3xh^2+h^3-x^3}{h}$

We can simplify this to

$f'(x)=\lim_{h\rightarrow0}\frac{3x^2h+3xh^2+h^3}{h}$

Now factor out an h to get

$f'(x)=\lim_{h\rightarrow0}\frac{h(3x^2+3xh+h^2)}{h}$

We can simplify and then evaluate the limit.

$f'(x)=\lim_{h\rightarrow0} 3x^2+3xh+h^2$

$\\ \\ =3x^2+3x(0)+0^2 \\ =3x^2$

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Example Question #5 : Definition Of Derivative

Use the definition of the derivative to solve for f'(x) .

f(x)=10x^2

Possible Answers:

20x

20x^2

Correct answer:

20x

Explanation:

In order to find f'(x) , we need to remember how to find f'(x) by using the definition of derivative.

Definition of Derivative:

$f'(x)=\lim_{h\rightarrow0} \frac{f(x+h)-f(x)}{h}$

Now lets apply this to our problem.

$f'(x)=\lim_{h\rightarrow0}\frac{10(x+h)^2-10x^2}{h}$

Now lets expand the numerator.

$f'(x)=\lim_{h\rightarrow0}\frac{10x^2+20xh+10h^2-10x^2}{h}$

We can simplify this to

$f'(x)=\lim_{h\rightarrow0}\frac{20xh+10h^2}{h}$

Now factor out an h to get

$f'(x)=\lim_{h\rightarrow0}\frac{h(20x+10h)}{h}$

We can simplify and then evaluate the limit.

$f'(x)=\lim_{h\rightarrow0} 20x+10h$

$\\ \\ =20x+10(0) \\ =20x$

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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1 : Derivatives

Example Question #2 : Derivatives

Example Question #3 : Derivatives

Example Question #4 : Definition Of Derivative

Example Question #4 : Definition Of Derivative

Example Question #5 : Definition Of Derivative

Given:

Find f'(x):

Example Question #4 : Definition Of Derivative

Example Question #3 : Derivatives

$f(x)=x^2\tan^{-1}(3x^5)$

Example Question #3 : Derivative Review

Example Question #5 : Definition Of Derivative

All Calculus 2 Resources

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