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Calculus 2 : Derivative Review

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Example Questions

Example Question #511 : Derivative Review

Find the derivative of the function:

$y=ln({\sqrt[3]{x}})+ln(\frac{1}{x})$

Possible Answers:

$y'=\frac{2x}{3}$

$y'=-\frac{2}{3x}$

$y'=\frac{2}{3x}$

$y'=\frac{1}{x^3}+x$

$y'=\frac{1}{3x}-x$

Correct answer:

$y'=-\frac{2}{3x}$

Explanation:

First we simplify the function using properties of logarithmic functions:

ln(x^a)=aln(x) and $ln(\frac{x}{y})=ln(x)-ln(y)$

Therefore:

$y=ln({\sqrt[3]x{}})+ln(\frac{1}{x})=\frac{1}{3}ln(x)+ln(1)-ln(x)$

also

ln(1)=0

Therefore

$y=\frac{1}{3}ln(x)-ln(x)$

and

$y'=\frac{1}{3}\frac{1}{x}-\frac{1}{x}=\frac{1-3}{3x}=-\frac{2}{3x}$

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Example Question #512 : Derivatives

Find the derivative of the function:

$y=cos(x)e^{5x}$

Possible Answers:

y'=e^x(5cos(x)-sin(x))

$y'=e^{5x}(cos(x)-sin(x))$

$y'=e^{5x}(5cos(x)+sin(x))$

$y'=e^{5x}(cos(x)+sin(x))$

$y'=e^{5x}(5cos(x)-sin(x))$

Correct answer:

$y'=e^{5x}(5cos(x)-sin(x))$

Explanation:

to derive this equation we use the product rule:

$(f\cdot g)'= f'\cdot g+g'\cdot f$

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

and

$g=e^{5x}$ $g=5e^{5x}$

Therefore:

$(cos(x)\cdot e^{5x})'=-sin(x\cdot e^{5x})+5cos(x)\cdot e^{5x}$

$=e^{5x}(5cos(x)-sin(x))$

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Example Question #512 : Derivative Review

Use logarithmic differentiation to compute the derivative of the function,

$\small \small y= \frac{\sqrt{2x^2-1}}{(x^3+1)(2x^2+1)^2}$

Possible Answers:

$\small \small \small \small y'=\frac{\sqrt{2x^2-1}}{(x^3+1)(2x^2+1)^2} \left(\frac{1}{\sqrt{2x^2-1}}-\frac{1}{x^3+1}-\frac{2}{2x^2+1} \right )$

$\small \small \small \small \small y'=\frac{1}{\sqrt{2x^2-1}}-\frac{1}{x^3+1}-\frac{2}{2x^2+1}$

$\small \small \small y'=\frac{\sqrt{2x^2-1}}{(x^3+1)(2x^2+1)^2} \left(\frac{4x}{\sqrt{2x^2-1}}-\frac{3x^2+1}{x^3+1}-\frac{8x}{2x^2+1} \right )$

$\small \small \small y'=\frac{\sqrt{2x^2-1}}{(x^3+1)(2x^2+1)^2} \left(\frac{4x^2+1}{\sqrt{2x^2-1}}-\frac{3x^3-1}{x^3+1}+\frac{8x^2}{2x^2+1} \right )$

$\small \small \small \small y'= \frac{4x}{\sqrt{2x^2-1}}-\frac{3x^2+1}{x^3+1}-\frac{8x}{2x^2+1}$

Correct answer:

$\small \small \small y'=\frac{\sqrt{2x^2-1}}{(x^3+1)(2x^2+1)^2} \left(\frac{4x}{\sqrt{2x^2-1}}-\frac{3x^2+1}{x^3+1}-\frac{8x}{2x^2+1} \right )$

Explanation:

$\small \small y= \frac{\sqrt{2x^2-1}}{(x^3+1)(2x^2+1)^2}$

Logarithmic differentiation exploits the properties of logarithms to easily compute derivatives for functions that would otherwise be extremely tedious to find. Direct differentiation using the quotient rule could become quite messy. Take the natural logarithm of both sides of the equation,

$\small \ln (y)= \ln\left(\frac{\sqrt{2x^2-1}}{(x^3+1)(2x^2+1)^2}\right)$ (1)

Expand the right-side using the properties of logarithms:

____________________________________________________________

Properties of Logarithmic Functions:

1. $\small \ln(ab)=\ln(a)+\ln(b)$

2. $\small \ln\left(\frac{a}{b}\right)=\ln(a)-\ln(b)$

3. $\small \ln(a^n)=n\ln(a)$

Then proceed with the differentiation using the known derivative of the natural logarithm function and the chain rule:

____________________________________________________________

Derivative of the Natural Logarithm

$\small \frac{d}{dx} \ln(x)= \frac{1}{x}$

For a function $\small y$ of $\small x$ , apply the chain-rule,

$\small \frac{d}{dx}\ln(y)-\frac{y'}{y}$

____________________________________________________________

Expanding the right-side of equation (1) first by using Property 2.

$\small \ln(y)=\ln\left(\sqrt{2x^2-1}\right )-\ln\left[(x^3+1)(2x^2+1)^2 \right ]$

Expand the second term with Property 1. Use Property 3 to pull out the exponent in the third term obtained after applying Property 1.

$\small \small \small \small \ln(y)=\ln\left(\sqrt{2x^2-1}\right )-\ln\left(x^3+1)-2\ln\left[\left(2x^2+1)\right\right]$

Differentiating implicitly over both sides of the equation with respect to $\small x$ . Be sure to apply the chain rule as needed.

$\small \small \small \frac{y'}{y}=\frac{1}{\sqrt{2x^2-1}}(4x)-\frac{1}{x^3+1}(3x^2+1)-\frac{2}{2x^2+1}(4x)$

So now the derivative we were looking for, $\small y'$ can be solved by multiplying both sides by $\small y$ and then substituting back in the original function to write everything in terms of $\small x.$

$\small \small \small y'=\frac{\sqrt{2x^2-1}}{(x^3+1)(2x^2+1)^2} \left(\frac{4x}{\sqrt{2x^2-1}}-\frac{3x^2+1}{x^3+1}-\frac{8x}{2x^2+1} \right )$

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Example Question #513 : Derivative Review

A farmer wants to fence off a piece of land that has a rectangular shape; he has 700 feet of fencing material. What is the maximum area he can fence off, given the amount of material he has?

Possible Answers:

49,000 ft^2

7000 ft^2

30,625 ft^2

30,000 ft^2

17,500 ft^2

Correct answer:

30,625 ft^2

Explanation:

The farmer's fencing material needs to cover the perimeter of his property. Since this piece of property is shaped like a rectangle, we know that the perimeter can be modeled with the equation

2l+2w=P .

In this case, we know that P=700 , since the 700 ft. of fencing need to fit around the whole property.

This problem wants to maximize the area, so we're trying to find which values maximize this equation:

A=lw .

We know that

2L+2w=700 , or simplified, that l+w=350 .

Solving for gives us

w=350-l ,

which we can plug into our area equation, giving us

A=l(350-l)

$A=350l-l^{2}$ .

Taking the first derivative gives us

A'=350-2l .

Making equal zero allows us to solve for .

0=350-2l .

So, is 175 ft. To determine if this value is a maximum length, or a minimum, we take the second derivative of our area equation, which yields a constant. Because this value is always less than zero, 175 ft. is a maximum. using our perimeter formula, we see that is also equal to 175 ft. So, the maximum area the farmer can fence off is 175 ft. x 175 ft., or 30,625 sq. ft.

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Example Question #513 : Derivatives

Differentiate the function:

y=x^3e^x

Possible Answers:

y'=x^2e^x(x+5)

y'=3x^2e^x

y'=3x^2e^x(x+2)

y'=3x^2+e^x

y'=x^2e^x(x-3)

Correct answer:

y'=x^2e^x(x+5)

Explanation:

on this problem we apply the product rule:

$(f\cdot g)'=f'\cdot g+f\cdot g' \,\!$

let: f=x^3 $f{}'=3x^2$ and g=e^x g'=e^x

$\therefore y'=(x^3 \cdot e^x)'= 3x^2\cdot e^x + e^x\cdot x^3$

=3x^2e^x+e^xx^3=x^2e^x(x+3)

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Example Question #512 : Derivative Review

Find the derivative of the function:

$y=sin(e^{3x})$

Possible Answers:

y'=3e^xcos(e^x)

$y'=e^{3x}sin(e^{3x})$

$y'=e^{3x}cos(e^{3x})$

$y'=3e^{3x}cos(e^{3x}})$

$y'=3e^{3x}sin(e^{3x})$

Correct answer:

$y'=3e^{3x}cos(e^{3x}})$

Explanation:

to find the derivative of this function we need to use the chain rule:

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

let

$u=e^{3x}$ and y=sin(u) $\therefore \frac{dy}{du}=cos(u)=cos(e^{3x})$

and

$\frac{du}{dx}=3e^{3x}$

and

$\frac{dy}{du}=3e^{3x}cos(e^3x)$ $\frac{dy}{dx}=3e^{3x}cos(e^{3x})$

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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #511 : Derivative Review

Example Question #512 : Derivatives

Example Question #512 : Derivative Review

Example Question #513 : Derivative Review

Example Question #513 : Derivatives

Example Question #512 : Derivative Review

All Calculus 2 Resources

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