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Calculus 2 : Calculus II

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 #11 : Definition Of Derivative

Use the limit definition of a derivative to find the derivative of the following function at x=3 :

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

Possible Answers:

Correct answer:

Explanation:

The limit definition of the derivative of a function is

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

where h represents a small change in x.

Now, using the given function, evaluate the limit:

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

which simplified becomes

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

Now, plug in the given x value into the derivative and we get .

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Example Question #12 : Definition Of Derivative

Find dy/dx by implicit differentiation:

$x^3+x\tan^{-1}(y)=e^y$

Possible Answers:

$\frac{3x^2-\tan^{-1}(y)}{e^y-\frac{x}{1+y^2}}$

$\frac{e^y-\frac{x}{1+y^2}}{3x^2+\tan^{-1}(y)}$

$\frac{e^y+\tan^{-1}(y)-x^3}{x}$

$\frac{3x^2+\tan^{-1}(y)}{e^y-\frac{x}{1+y^2}}$

$\frac{e^y+3x^2}{\tan^{-1}(y)-\frac{x}{1+y^2}}$

Correct answer:

$\frac{3x^2+\tan^{-1}(y)}{e^y-\frac{x}{1+y^2}}$

Explanation:

To find dy/dx we must take the derivative of the given function implicitly. Notice the term $x\tan^{-1}(y)$ will require the use of the Product Rule, because it is a composition of two separate functions multiplied by each other. Every other term in the given function can be derived in a straight-forward manner, but this term tends to mess with many students. Remember to use the Product Rule:

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

Now if we take the derivative of each component of the given problem statement:

$3x^2\frac{dx}{dx}+[(x)(\frac{1}{1+y^2}(\frac{dy}{dx}))+(1)(\frac{dx}{dx})(\tan^{-1}(y))]=e^y\frac{dy}{dx}$

Notice that anytime we take the derivative of a term with x involved we place a "dx/dx" next to it, but this is equal to "1".

So this now becomes:
$3x^2+[\frac{x}{1+y^2}\frac{dy}{dx}+\tan^{-1}(y)}]=e^y\frac{dy}{dx}$

Now if we place all the terms with a "dy/dx" onto one side and factor out we can solved for it:

$3x^2+\tan^{-1}(y)=e^y\frac{dy}{dx}-\frac{x}{1+y^2}\frac{dy}{dx}$

$3x^2+\tan^{-1}(y)=\frac{dy}{dx}(e^y-\frac{x}{1+y^2})$

$\frac{dy}{dx}=\frac{3x^2+\tan^{-1}(y)}{e^y-\frac{x}{1+y^2}}$

This is one of the answer choices.

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Example Question #13 : Definition Of Derivative

Find dx/dy by implicit differentiation:

$x^3+x\tan^{-1}(y)=e^y$

Possible Answers:

$\frac{3x^2+e^y}{\tan^{-1}(y)-\frac{x}{1+y^2}}$

$\frac{\frac{x}{1+y^2}+e^y}{5x^2-\tan^{-1}(y)}$

$\frac{e^y-\frac{x}{1+y^2}}{3x^2+\tan^{-1}(y)}$

$\frac{3x^2+\tan^{-1}(y)}{e^y-\frac{x}{1+y^2}}$

$\frac{\tan^{-1}(y)-\frac{x}{1+y^2}}{3x^2+e^y}$

Correct answer:

$\frac{e^y-\frac{x}{1+y^2}}{3x^2+\tan^{-1}(y)}$

Explanation:

To find dx/dy we must take the derivative of the given function implicitly. Notice the term $x\tan^{-1}(y)$ will require the use of the Product Rule, because it is a composition of two separate functions multiplied by each other. Every other term in the given function can be derived in a straight-forward manner, but this term tends to mess with many students. Remember to use the Product Rule:

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

Now if we take the derivative of each component of the given problem statement:

$3x^2\frac{dx}{dy}+[(x)(\frac{1}{1+y^2}(\frac{dy}{dy}))+(1)(\frac{dx}{dy})(\tan^{-1}(y))]=e^y\frac{dy}{dy}$

Notice that anytime we take the derivative of a term with y involved we place a "dy/dy" next to it, but this is equal to "1".

So this now becomes:
$3x^2\frac{dx}{dy}+[\frac{x}{1+y^2}+\tan^{-1}(y)\frac{dx}{dy}]=e^y$

Now if we place all the terms with a "dx/dy" onto one side and factor out we can solved for it:

$3x^2\frac{dx}{dy}+\tan^{-1}(y)\frac{dx}{dy}=e^y-\frac{x}{1+y^2}$

$\frac{dx}{dy}[3x^2+\tan^{-1}(y)]=e^y-\frac{x}{1+y^2}$

$\frac{dx}{dy}=\frac{e^y-\frac{x}{1+y^2}}{3x^2+\tan^{-1}(y)}$

This is one of the answer choices.

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Example Question #17 : Derivative Review

Find the first derivative of the given function

f(x)=sin (x) .

Possible Answers:

f'(x)=cos (x)

f'(x)=tan (x)

f'(x)=-sin (x)

f'(x)=-cos (x)

Correct answer:

f'(x)=cos (x)

Explanation:

In order to find the first derivative

f'(x)

we must derive both sides of the equation since

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}f(x)$

$=\frac{\mathrm{d} }{\mathrm{d} x}[sin(x)]$

From the definition of the derivative of the sine function we have

$\frac{\mathrm{d} }{\mathrm{d} x}[sin(x)]=cos(x)$

As such, we have

f'(x)=cos (x)

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Example Question #18 : Derivative Review

Find the derivative of the following function using the limit definition:

f(x)=5x^2+x

Possible Answers:

10x+xe^x

5x+1

10x+1

Correct answer:

10x+1

Explanation:

The limit definition of a derivative is

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

where h represents a small change in x.

Now, use the above formula for the given function:

$\lim_{h\rightarrow 0}\left[\frac{5(x+h)^2+(x+h)-(5x^2+x)}{h}\right]$

which simplified becomes

$\lim_{h\rightarrow 0}h\frac{(10x+1)}{h}=10x+1$ .

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Example Question #21 : Definition Of Derivative

Using the limit definition of a derivative, find the velocity function of a particle if its position is given by:

s(t)=4t^2+t+3

Possible Answers:

8t+3

8t+1

Correct answer:

8t+1

Explanation:

The limit definition of a derivative is

$\lim_{h\rightarrow 0}\left[\frac{f(x+h)-f(x)}{h}\right]$

where h represents a very small change in x.

Because the velocity function is simply the first derivative of the position function, we can use the above formula, with the position function as f(x), to find the velocity function:

$\lim_{h\rightarrow 0}\frac{4(t+h)^2+(t+h)+3-(4t^2+t+3)}{h}$

which simplified becomes

$\lim_{h\rightarrow 0}\frac{h(8t+1)}{h}=8t+1=v(t)$

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Example Question #1141 : Calculus Ii

What is the derivative of $f(x)=x^{200}\sin x$ ?

Possible Answers:

$f'(x)=x^{199}(200\cos x+x\sin x)$

$f'(x)=x^{199}(200\sin x+x\cos x)$

$f'(x)=200x^{199}\cos x$

$f'(x)=-\frac{x^{201}}{201}\cos x$

Correct answer:

$f'(x)=x^{199}(200\sin x+x\cos x)$

Explanation:

The derivative of

$f(x)=x^{200}\sin x$

is found using the chain rule:

(h(x)g(x))'=h'(x)g(x)+h(x)g'(x)

So we have

$f'(x)=200x^{199}\sin x+x^{200}\cos x=x^{199}(200\sin x+x\cos x)$

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Example Question #23 : Definition Of Derivative

What is the derivative of

$f(x)=(\sin x^2+x^5-x)^7$ ?

Possible Answers:

$f'(x)=7(-2x\cos x^2+5x^4-1)(\sin x^2+x^5-x)^6$

$f'(x)=(2x\cos x^2+5x^4-1)(\sin x^2+x^5-x)^7$

$f'(x)=(2x\cos x^2+5x^4-1)(\sin x^2+x^5-x)^6$

$f'(x)=7(2x\cos x^2+5x^4-1)(\sin x^2+x^5-x)^6$

Correct answer:

$f'(x)=7(2x\cos x^2+5x^4-1)(\sin x^2+x^5-x)^6$

Explanation:

To find the derivative of $f(x)=(\sin x^2+x^5-x)^7$ , we simply use the chain rule, which is

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

So then we have

$f'(x)=7(2x\cos x^2+5x^4-1)(\sin x^2+x^5-x)^6$

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Example Question #24 : Definition Of Derivative

Given f(x)=5x^2+2 , what is f'(10) ?

Possible Answers:

f'(10)=502

f'(10)=100

f'(10)=2

f'(10)=25

Correct answer:

f'(10)=100

Explanation:

We can find the derivative of f(x)=5x^2+2 and simply plug in x=10 to get f'(10) :

To find the derivative we will need to use the power rule which states,

$x^n\ dx\rightarrow nx^{n-1}$ .

Also recall that the derivative of a constant is zero.

Applying the power rule we get the following.

$f'(x)=2\cdot 5x^{2-1}$

f'(x)=10x

so then we get

$f'(10)=10\cdot 10=100$ .

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Example Question #25 : Definition Of Derivative

Given $f(x)=\sin x^2$ , what is $f'\left(\sqrt{\frac{\pi}{2}}\right)$ ?

Possible Answers:

$f'\left(\sqrt{\frac{\pi}{2}}\right)=\sqrt{\pi}$

$f'\left(\sqrt{\frac{\pi}{2}}\right)=0$

$f'\left(\sqrt{\frac{\pi}{2}}\right)=2\sqrt{\frac{\pi}{2}}$

$f'\left(\sqrt{\frac{\pi}{2}}\right)=\sqrt{\frac{\pi}{2}}$

Correct answer:

$f'\left(\sqrt{\frac{\pi}{2}}\right)=0$

Explanation:

We can find the derivative of $f(x)=\sin x^2$ and simply plug in $x=\sqrt{\frac{\pi}{2}}$ to get $f'\left(\sqrt{\frac{\pi}{2}}\right)$ :

To find the derivative of this function we will need to use the chain rule which states,

$f(g(x))\ dx\rightarrow f'(g(x))\cdot g'(x)$

the power rule which states,

$x^n\ dx\rightarrow nx^{n-1}$ .

Also recall that the derivative of sine is cosine.

Applying these rules we get the following derivative.

$f'(x)=\cos x^2\cdot 2x^{2-1}$

$f'(x)=2x\cos x^2$

so then we get

$f'\left(\sqrt{\frac{\pi}{2}}\right)=2\sqrt{\frac{\pi}{2}}\cos \frac{\pi}{2}=2\sqrt{\frac{\pi}{2}}\cdot 0=0$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #11 : Definition Of Derivative

Example Question #12 : Definition Of Derivative

Find dy/dx by implicit differentiation:

Example Question #13 : Definition Of Derivative

Find dx/dy by implicit differentiation:

Example Question #17 : Derivative Review

Example Question #18 : Derivative Review

Example Question #21 : Definition Of Derivative

Example Question #1141 : Calculus Ii

Example Question #23 : Definition Of Derivative

Example Question #24 : Definition Of Derivative

Example Question #25 : Definition Of Derivative

All Calculus 2 Resources

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