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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 #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 #41 : Derivative Review

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 #1171 : Calculus Ii

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:

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

None of the other answers

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

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

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

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 #41 : Derivative Review

Possible Answers:

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

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

f'(x)=2xcos(x)

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

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

f''(x)=-x^2sin(x)+4xcos(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 #41 : Definition Of Derivative

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

Possible Answers:

f'(x)=10

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

f'(x)=5x+12

f'(x)=10x+12

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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Example Question #51 : Definition Of Derivative

Find the derivative of f(x)=sinx using the definition of the derivative.

Possible Answers:

f'(x)=-cos(x)

f'(x)=-sin(x)

f'(x)=sin(x)

f'(x)=cos(x)

Correct answer:

f'(x)=cos(x)

Explanation:

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

f(x)=sinx

f(x+h)=sin(x+h)

Substituting these expressions into the definition of a derivative gives us

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

Substituting in the trigonometric identity

sin(A+B)=sinAcosB+cosAsinB

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

$f'(x)=\lim_{h\rightarrow 0}\frac{sin(x)(cos(h)-1)+cos(x)sin(h)}{h}$

Then using the identities

$\lim_{h\rightarrow 0}\frac{cos(h)-1}{h}=0$ and $\lim_{h\rightarrow 0}\frac{sin(h)}{h}=1$

$f'(x)=\lim_{h\rightarrow 0}\left [sin(x)\frac{(cos(h)-1)}{h}+cos(x)\frac{sin(h)}{h} \right ]$

$f'(x)=sin(x)\lim_{h\rightarrow 0}\left [\frac{(cos(h)-1)}{h} \right ]+cos(x)\lim_{h\rightarrow 0}\left [\frac{sin(h)}{h} \right ]$

f'(x)=sin(x)*0+cos(x)*1

f'(x)=cos(x)

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Example Question #52 : Derivative Review

Find the derivative of f(x)=cosx using the definition of the derivative.

Possible Answers:

f'(x)=-cos(x)

f'(x)=-sin(x)

f'(x)=sin(x)

f'(x)=cos(x)

Correct answer:

f'(x)=sin(x)

Explanation:

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

f(x)=cosx

f(x+h)=cos(x+h)

Substituting these expressions into the definition of a derivative gives us

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

Substituting in the trigonometric identity

cos(A-B)=cosAcosB+sinAsinB

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

$f'(x)=\lim_{h\rightarrow 0}\frac{cos(x)(cos(h)-1)+sin(x)sin(h)}{h}$

Then using the identities

$\lim_{h\rightarrow 0}\frac{cos(h)-1}{h}=0$ and $\lim_{h\rightarrow 0}\frac{sin(h)}{h}=1$

$f'(x)=\lim_{h\rightarrow 0}\left [cos(x)\frac{(cos(h)-1)}{h}+sin(x)\frac{sin(h)}{h} \right ]$

$f'(x)=cos(x)\lim_{h\rightarrow 0}\left [\frac{(cos(h)-1)}{h} \right ]+sin(x)\lim_{h\rightarrow 0}\left [\frac{sin(h)}{h} \right ]$

f'(x)=cos(x)*0+sin(x)*1

f'(x)=sin(x)

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Example Question #53 : Derivative Review

Find the derivative of $f(x)=\frac{x+3}{x-2}$ using the definition of the derivative.

Possible Answers:

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

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

$f'(x)=\frac{-5}{x^{2}-4x+4}$

f'(x)=1

Correct answer:

$f'(x)=\frac{-5}{x^{2}-4x+4}$

Explanation:

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

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

$f(x+h)=\frac{x+h+3}{x+h-2}$

Substituting these expressions into the definition of a derivative gives us

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

$f'(x)=\lim_{h\rightarrow 0}\frac{1}{h}*\left (\frac{x+h+3}{x+h-2}-\frac{x+3}{x-2} \right )$

$f'(x)=\lim_{h\rightarrow 0}\frac{1}{h}*\left (\frac{(x+h+3)(x-2)-(x+3)(x+h-2)}{(x+h-2)(x-2)}\right )$

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

$f'(x)=\lim_{h\rightarrow 0} \frac{(x^{2}+hx+3x-2x-2h-6)-(x^{2}+hx-2x+3x+3h-6)}{h(x^{2}+hx-2x-2x-2h+4)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{x^{2}+hx+x-2h-6-x^{2}-hx-x-3h+6}{h(x^{2}+hx-4x-2h+4)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{-5h}{h(x^{2}+hx-4x-2h+4)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{-5}{(x^{2}+hx-4x-2h+4)}$

$f'(x)=\frac{-5}{x^{2}+(0)x-4x-2(0)+4}$

$f'(x)=\frac{-5}{x^{2}-4x+4}$

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Example Question #54 : Derivative Review

Find the derivative of $f(x)=\frac{5x^2-1}{x+3}$ using the definition of the derivative.

Possible Answers:

f'(x)={10x}

$f'(x)=\frac{5x^2+30x+1}{x^2+6x+9}$

$f'(x)=\frac{x^2+6x-2}{x^2+3x-7}$

$f'(x)=\frac{5x^2-1}{x+3}$

Correct answer:

$f'(x)=\frac{5x^2+30x+1}{x^2+6x+9}$

Explanation:

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

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

$f(x+h)=\frac{5(x+h)^2-1}{x+h+3}$

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{\frac{5(x+h)^2-1}{x+h+3}-\frac{5(x)^2-1}{x+3}}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{1}{h}*\left (\frac{5(x+h)^2-1}{x+h+3}-\frac{5(x)^2-1}{x+3} \right )$

$f'(x)=\lim_{h\rightarrow 0} \frac{(5(x+h)^{2}-1)(x+3)-(5x^{2}-1)(x+h+3)}{h(x+h+3)(x+3)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{(5(x^2+2xh+h^2)-1)(x+3)-(5x^{2}-1)(x+h+3)}{h(x+h+3)(x+3)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{(5x^2+10xh+5h^2-1)(x+3)-(5x^{2}-1)(x+h+3)}{h(x+h+3)(x+3)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{(5x^3+10x^2h+5h^2x-x+15x^2+30xh+15h^2-3)-(5x^3+5hx^2+15x^2-x-h-3)}{h(x^2+hx+3x+3x+3h+9)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{5x^3+10x^2h+5h^2x-x+15x^2+30xh+15h^2-3-5x^3-5hx^2-15x^2+x+h+3}{h(x^2+hx+3x+3x+3h+9)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{5x^2h+5h^2x+30xh+15h^2+h}{h(x^2+hx+6x+3h+9)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{h(5x^2+5hx+30x+15h+1)}{h(x^2+hx+6x+3h+9)}$

$f'(x)=\lim_{h\rightarrow 0} \frac{5x^2+5hx+30x+15h+1}{x^2+hx+6x+3h+9}$

$f'(x)=\frac{5x^2+5(0)x+30x+15(0)+1}{x^2+(0)x+6x+3(0)+9}$

$f'(x)=\frac{5x^2+30x+1}{x^2+6x+9}$

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Example Question #51 : Definition Of Derivative

Find the derivative of $f(x)=x^{3}-3x^{2}-6x+10$ using the definition of the derivative.

Possible Answers:

f(x)=10

$f'(x)=3x^{3}-x^{2}-4$

f'(x)=3x^2-6x-6

$f'(x)=x^{3}-3x^{2}-6x+10$

Correct answer:

f'(x)=3x^2-6x-6

Explanation:

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

$f(x)=x^{3}-3x^{2}-6x+10$

$f(x+h)=(x+h)^{3}-3(x+h)^{2}-6(x+h)+10$

f(x+h)=(x^3+3xh^2+3x^2h+h^3)-3(x^2+2xh+h^2)-6(x+h)+10

f(x+h)=x^3+3xh^2+3x^2h+h^3-3x^2-6xh-3h^2-6x-6h+10

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{x^3+3xh^2+3x^2h+h^3-3x^2-6xh-3h^2-6x-6h+10-(x^{3}-3x^{2}-6x+10)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{x^3+3xh^2+3x^2h+h^3-3x^2-6xh-3h^2-6x-6h+10-x^{3}+3x^{2}+6x-10}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{3xh^2+3x^2h+h^3-6xh-3h^2-6h}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{h(3xh+3x^2+h^2-6x-3h-6)}{h}$

$f'(x)=\lim_{h\rightarrow 0}3xh+3x^2+h^2-6x-3h-6$

f'(x)=3x(0)+3x^2+(0)^2-6x-3(0)-6

f'(x)=3x^2-6x-6

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #46 : Derivatives

Find dy/dx:

Example Question #41 : Derivative Review

Example Question #1171 : Calculus Ii

Example Question #41 : Derivative Review

Example Question #41 : Definition Of Derivative

Example Question #51 : Definition Of Derivative

Example Question #52 : Derivative Review

Example Question #53 : Derivative Review

Example Question #54 : Derivative Review

Example Question #51 : Definition Of Derivative

All Calculus 2 Resources

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