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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 #51 : Derivative Review

Find the derivative of $f(x)=\sqrt{x^{2}-4x}$ using the definition of the derivative.

Possible Answers:

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

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

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

$f'(x)=\sqrt{2x-4}$

Correct answer:

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

Explanation:

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

$f(x)=\sqrt{x^{2}-4x}$

$f(x+h)=\sqrt{(x+h)^{2}-4(x+h)}$

Substituting these expressions into the definition of a derivative gives us

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

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

Next we will multiply the numerator and denominator by the complex conjugate of the numerator

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

$f'(x)=\lim_{h\rightarrow 0}\frac{\left (\sqrt{x^2+2xh+h^2-4x-4h} \right )^2-\left (\sqrt{x^{2}-4x} \right )^2}{h*(\sqrt{x^2+2xh+h^2-4x-4h}+\sqrt{x^{2}-4x})}$

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

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

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

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

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

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

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

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

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Example Question #51 : Definition Of Derivative

Find the derivative of f(x)=(x-7)(x-9) using the definition of the derivative.

Possible Answers:

f'(x)=x^2

f'(x)=2x-16

f'(x)=5x-2

f'(x)=7

Correct answer:

f'(x)=2x-16

Explanation:

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

f(x)=(x-7)(x-9)

f(x)=x^2-7x-9x+63

f(x)=x^2-16x+63

f(x+h)=(x+h-7)(x+h-9)

f(x+h)=(x^2+hx-7x+hx+h^2-7h-9x-9h+63)

f(x+h)=(x^2+2hx-16x+h^2-16h+63)

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{(x^2+2hx-16x+h^2-16h+63)-(x^2-16x+63)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{x^2+2hx-16x+h^2-16h+63-x^2+16x-63}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{2hx+h^2-16h}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{h*(2x+h-16)}{h}$

$f'(x)=\lim_{h\rightarrow 0}(2x+h-16)$

f'(x)=2x+0-16

f'(x)=2x-16

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Example Question #58 : Derivative Review

Find the derivative of $f(x)=\sqrt{x-7}$ using the definition of the derivative.

Possible Answers:

$f'(x)=\sqrt{x-7}$

$f'(x)=\frac{1}{2\sqrt{x-7}}$

f(x)=1

$f'(x)=\frac{x-7}{\sqrt{x-7}}$

Correct answer:

$f'(x)=\frac{1}{2\sqrt{x-7}}$

Explanation:

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

$f(x)=\sqrt{x-7}$

$f(x+h)=\sqrt{x+h-7}$

Substituting these expressions into the definition of a derivative gives us

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

Next we will multiply the numerator and denominator by the complex conjugate of the numerator

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

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

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

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

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

$f'(x)=\lim_{h\rightarrow 0}\frac{1}{\sqrt{x+h-7}+\sqrt{x-7}}$

$f'(x)=\frac{1}{\sqrt{x+(0)-7}+\sqrt{x-7}}$

$f'(x)=\frac{1}{2\sqrt{x-7}}$

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Example Question #59 : Derivative Review

Find the derivative of f(x)=(2x+3)(x-16) using the definition of the derivative.

Possible Answers:

f'(x)=1

f'(x)=2x^2

f'(x)=4x-29

f'(x)=3x-16

Correct answer:

f'(x)=4x-29

Explanation:

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

f(x)=(2x+3)(x-16)

f(x)=2x^2+3x-32x-48

f(x)=2x^2-29x-48

f(x+h)=(2(x+h)+3)(x+h-16)

f(x+h)=(2x+2h+3)(x+h-16)

f(x+h)=(2x^2+2hx+3x+2xh+2h^2+3h-32x-32h-48)

f(x+h)=2x^2+2h^2+4hx-29x-29h-48

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{(2x^2+2h^2+4hx-29x-29h-48)-(2x^2-29x-48)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{2x^2+2h^2+4hx-29x-29h-48-2x^2+29x+48}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{2h^2+4hx-29h}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{h(2h+4x-29)}{h}$

$f'(x)=\lim_{h\rightarrow 0}(2h+4x-29)$

f'(x)=(2(0)+4x-29)

f'(x)=4x-29

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Example Question #60 : Derivative Review

Find the derivative of f(x)=(x+2)(x+3)(x+4) using the definition of the derivative.

Possible Answers:

f'(x)=-x^2+7x +37

f'(x)=4x^2+7x +49

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

f'(x)=3x^2+18x +26

Correct answer:

f'(x)=3x^2+18x +26

Explanation:

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

f(x)=(x+2)(x+3)(x+4)

f(x)=(x^2+5x+6)(x+4)

f(x)=x^3+5x^2+6x +4x^2+20x+24

f(x)=x^3+9x^2+26x+24

f(x+h)=(x+h+2)(x+h+3)(x+h+4)

f(x+h)=(x^2+hx+2x +hx+h^2+2h +3x+3h+6)(x+h+4)

f(x+h)=(x^2+2hx+5x+h^2+5h+6)(x+h+4)

f(x+h)=(x^3+2hx^2+5x^2+h^2x+5hx+6x +x^2h+2h^2x+5xh+h^3+5h^2+6h +4x^2+8hx+20x+4h^2+20h+24)

f(x+h)=(x^3+h^3+3x^2h+3h^2x+9x^2+9h^2+18hx +26x+26h +24)

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{(x^3+h^3+3x^2h+3h^2x+9x^2+9h^2+18hx +26x+26h +24)-(x^3+9x^2+26x+24)}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{x^3+h^3+3x^2h+3h^2x+9x^2+9h^2+18hx +26x+26h +24-x^3-9x^2-26x-24}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{h^3+3x^2h+3h^2x+9h^2+18hx +26h}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{h*(h^2+3x^2+3hx+9h+18x +26)}{h}$

$f'(x)=\lim_{h\rightarrow 0}(h^2+3x^2+3hx+9h+18x +26)$

f'(x)=(0)^2+3x^2+3(0)x+9(0)+18x +26

f'(x)=3x^2+18x +26

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Example Question #61 : Derivatives

Find the derivative of $f(x)=\frac{x}{x+1}$ using the definition of the derivative.

Possible Answers:

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

f'(x)=1

$f'(x)=\frac{1}{2x+1}$

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

Correct answer:

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

Explanation:

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

$f(x)=\frac{x}{x+1}$

$f(x+h)=\frac{x+h}{x+h+1}$

Substituting these expressions into the definition of a derivative gives us

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

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

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

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

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

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

$f'(x)=\left (\frac{1}{x^2+(0)x+2x+(0)+1} \right )$

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

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Example Question #1181 : Calculus Ii

Possible Answers:

4x+C

x^2

Correct answer:

Explanation:

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Example Question #1182 : Calculus Ii

What is the derivative of a term: ax^2 ?

Possible Answers:

2x^3

2ax

3x^2

Correct answer:

2ax

Explanation:

Step 1: Move the exponent to the coefficient..

We get, 2ax..

Step 2: The new exponent after moving the old one down is less than it was before:

$2ax^{2-1}=2ax$

The derivative of ax^2 is 2ax

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Example Question #64 : Derivatives

Find the function represented by the following limit,

$\lim_{ h->0}\frac{\ln[(y+h)^2]-\ln(h^2)}{h}$

Possible Answers:

y^2 -1

$\frac{1}{y^2}+1$

$\frac{2}{y}$

$\ln(y+4)$

Constant

Correct answer:

$\frac{2}{y}$

Explanation:

$\lim_{ h->0}\frac{\ln[(y+h)^2]-\ln(h^2)}{h}$ (1)

Recall the first principles definition of a derivative,

$f'(y) = \lim_{h->0}\frac{f(y+h)-f(h)}{h}$ (2)

The limit in Equation (1) can be determined by inspection if we compare to the general definition of a derivative, Equation (2). It's apparent that equation (1) is the definition of the derivative for the function $\ln(y^2)$ . So the limit can be found by simply differentiating the function $\ln(y^2)$ with respect to .

$\frac{d}{dy}(\ln(y^2))=(2y)\frac{1}{y^2}=\frac{2}{y}$

$\lim_{ h->0}\frac{\ln[(y+h)^2]-\ln(h^2)}{h}=\frac{2}{y}$

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Example Question #65 : Derivatives

Find the derivative of the following function:

$f(x)=\frac{cosx}{sinx}$ .

Possible Answers:

-csc^2x

csc^2x

secxtanx

sec^2x

-sec^2x

Correct answer:

-csc^2x

Explanation:

The function in the problem is simply cotx ! Therefore, by knowing the derivative of cotx , we know the solution is -csc^2x . You would obtain the same results if you completed this problem using the quotient rule and simplifying.

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #51 : Derivative Review

Example Question #51 : Definition Of Derivative

Example Question #58 : Derivative Review

Example Question #59 : Derivative Review

Example Question #60 : Derivative Review

Example Question #61 : Derivatives

Example Question #1181 : Calculus Ii

Example Question #1182 : Calculus Ii

Example Question #64 : Derivatives

Example Question #65 : Derivatives

All Calculus 2 Resources

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