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Calculus 1 : Other Differential Functions

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Example Questions

Example Question #151 : How To Find Differential Functions

Differentiate the function:

f(x)=(3x+1)^2

Possible Answers:

f'(x)=18x+6

f'(x)=9x^2+6x+1

f'(x)=6

f'(x)=9

Correct answer:

f'(x)=18x+6

Explanation:

Take away the parentheses by multiplying the binomial by itself so that you are left with a polynomial.

f(x)=(3x+1)^2=9x^2+6x+1

Then take the derivative of each term in the polynomial using the power rule $nx^{n-1}$ where n is the exponent:

$\\f'(x)=\frac{d}{dx}(9x^2+6x+1)\\ \\=\frac{d}{dx}(9x^2)+\frac{d}{dx}(6x)+\frac{d}{dx}(1)\\ \\=2\cdot 9x^{2-1}+1\cdot 6x^{^{1-1}}\\ \\=18x+6$

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Example Question #334 : Differential Functions

Differentiate the function:

$f(x)=\sqrt{x}(x-1)$

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

Take away the parentheses by multiplying the outside piece to each term within the binomial so that you are left with a polynomial.

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

Then take the derivative of each term in the polynomial using the power rule $nx^{n-1}$ where n is the exponent:

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

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Example Question #151 : How To Find Differential Functions

Differentiate the function:

$f(x)=\frac{x^2+3x+4}{\sqrt{x}}$

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

Bring the root to the numerator. Take away the parentheses by multiplying the outside piece by each of the terms in the parentheses so that you are left with a polynomial.

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

Then take the derivative of each term in the polynomial using the power rule $nx^{n-1}$ where n is the exponent:

$\\ f'(x)=\frac{d}{dx}(x^{\frac{3}{2}}+3x^{\frac{1}{2}}+4x^{-\frac{1}{2}})\\ \\=\frac{3}{2}x^{\frac{3}{2}-1}+\frac{1}{2}\cdot 3x^{\frac{1}{2}-1}+\left(-\frac{1}{2}\cdot 4x^{-\frac{1}{2}-1}\right)\\ \\=\frac{3}{2}x^{\frac{1}{2}}+\frac{3}{2}x^{-\frac{1}{2}}-2x^{-\frac{3}{2}}$

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Example Question #151 : How To Find Differential Functions

Differentiate the function:

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

Possible Answers:

$f'(x)=x\sqrt{5}$

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

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

$f'(x)=\sqrt{5x}$

Correct answer:

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

Explanation:

Take the derivative of each term in the polynomial using the power rule $nx^{n-1}$ where n is the exponent:

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

$\\f'(x)=\frac{d}{dx}(x\sqrt{3}+x^{\frac{1}{2}}\sqrt{2})\\ \\=1\cdot x^{1-1}\cdot \sqrt{3}+\frac{1}{2}x^{\frac{1}{2}-1}\cdot \sqrt{2}\\ \\=\sqrt{3}+\frac{1}{2}\cdot \sqrt{2}x^{-\frac{1}{2}}\\ \\=\sqrt{3}+x^{-\frac{1}{2}}\cdot \frac{\sqrt{2}}{2}$

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Example Question #151 : Other Differential Functions

Differentiate the function:

$f(x)=3x\left(\frac{1}{x}+x^2\right)$

Possible Answers:

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

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

f'(x)=9x^2

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

Correct answer:

f'(x)=9x^2

Explanation:

Take away the parentheses by multiplying the outside piece by each term within the parentheses so that you are left with a polynomial.

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

Then take the derivative of each term in the polynomial using the power rule $nx^{n-1}$ where n is the exponent:

$\\f'(x)=\frac{d}{dx}(3+3x^3)\\ \\=\frac{d}{dx}(3)+\frac{d}{dx}(3x^3)\\ \\=0+3\cdot x^{3-1}\\ \\=9x^2$

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Example Question #341 : Differential Functions

A company wants to find the smallest amount of material to build a box with a square bottom of volume 6,250 mm^3 . Find the height of such a box.

Possible Answers:

None of the above.

25mm

50mm

11.61mm

0.25mm

Correct answer:

11.61mm

Explanation:

To solve this problem, we must first know that the surface area of a box with a square bottom would be A=4sh+s^2 where is the length of the side of the box and is the height of the box.

The volume of the box can be found with the formula V=s^2h where is the length of the side of the box and is the height of the box.

We know that the volume of such a box must be equivalent to 6,250; therefore to solve this problem we must first solve one of the variables in terms of another.

$\frac{6,250}{s^2}=h$

Then we must plug it into the surface area equation.

$A=\frac{(4)(6,250)}{s}+s^2$

Then, take the derivative and set it equal to zero to find the minimum. To take the derivative of that problem, one must know the power rule

$\frac{d}{dx}x^n=nx^{n-1}$ .

$0=-\frac{(4)(6,250)}{s^2}+2s$

Solving for

2s^3=25,000

s^3=12,500

s=23.2 mm

Plugging back into the original volume equation to solve for .

6,250=(23.2)^2h

h=11.61mm

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Example Question #342 : Differential Functions

Differentiate the equation.

$y=\frac{ln(x)sin(2x)}{x^4}$

Possible Answers:

$y'=\frac{x^4(ln(x)2cos(2x)+\frac{sin(2x)}{x})-4x^3ln(x)sin(2x)}{x^8}$

$y'=\frac{x^4(ln(x)2cos(x)+\frac{sin(2x)}{x})-4x^3ln(x)sin(2x)}{x^8}$

$y'=\frac{x^4(ln(x)2sin(2x)+\frac{sin(2x)}{x})-4x^3ln(x)sin(2x)}{x^8}$

$y'=\frac{x^4(ln(x)2cos(2x)+\frac{sin(2x)}{x})+4x^3ln(x)sin(2x)}{x^4}$

None of the above.

Correct answer:

$y'=\frac{x^4(ln(x)2cos(2x)+\frac{sin(2x)}{x})-4x^3ln(x)sin(2x)}{x^8}$

Explanation:

To approach this problem you msut first understand the different rules of calculus.

First is the Quotient rule,

$\frac{d}{dx}\frac{u}{v}=\frac{u'v-uv'}{v^2}$ .

Second is the product rule,

$\frac{d}{dx}uv=u'v+uv'$ .

Third is the power rule,

$\frac{d}{dx}x^n=nx^{n-1}$ .

Fourth is the chain rule,

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

Lastly, some identities must be known these are $\frac{d}{dx}sinx=cosx$ and $\frac{d}{dx}ln(x)=\frac{1}{x}$ .

The step by step solution of the equation above is as follows

First we apply the quotient rule and power rule.

$y'=\frac{x^4(ln(x)sin(2x))'-4x^3ln(x)sin(2x)}{x^8}$

Next the product rule and chain rule as well as the identities to take the derivative of ln(x)sin(2x) , giving our final answer.

$y'=\frac{x^4(ln(x)2cos(2x)+\frac{sin(2x)}{x})-4x^3ln(x)sin(2x)}{x^8}$

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Example Question #152 : How To Find Differential Functions

Given that $y=sin(x)+e^{2x}$ find $y^{(11)}$

Possible Answers:

$-cos(x)+e^{2x}$

$sin(x)+2^{11}e^{2x}$

$cos(x)+2^{11}e^{2x}$

$-sin(x)+2^{11}e^{2x}$

$-cos(x)+2^{11}e^{2x}$

Correct answer:

$-cos(x)+2^{11}e^{2x}$

Explanation:

The first few derivatives can be evaluated to give a general trend:

$y=sin(x)+e^{2x}$

$y'=cos(x)+2e^{2x}$

$y''=-sin(x)+4e^{2x}$

$y^{(3)}=-cos(x)+8e^{2x}$

$y^{(4)}=cos(x)+16e^{2x}$

Thus:

$n=1,5,9,13,..., y^{(n)}=cos(x)+2^ne^{2x}$

$n=2,6,10,14,..., y^{(n)}=-sin(x)+2^ne^{2x}$

$n=3,7,11,15,..., y^{(n)}=-cos(x)+2^ne^{2x}$

$n=4,8,12,16,..., y^{(n)}=sin(x)+2^ne^{2x}$

Therefore when n=11 :

$y^{(11)}=-cos(x)+2^{11}e^{2x}$

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Example Question #341 : Functions

Differentiate:

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

Possible Answers:

f'(x)=(3x^2+4x)^5

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

f'(x)=(36x+24)(3x^2+4x)^5

f'(x)=(36x+24)(3x^2+4x)

f'(x)=(36x+24)(3x^2+4x)^6

Correct answer:

f'(x)=(36x+24)(3x^2+4x)^5

Explanation:

To differentiate this function, you must use the chain rule. To do that, you first multiply the exponent by the coefficient in front of the expression. In this case, it's one. Then, you subtract one from the exponent, so at this point, you have 6(3x^2+4x)^5 .

The last step is to take the derivative of what's inside the parantheses, which is 6x+4 .

Multiply like terms so that your derivative is

f'(x)=(36x+24)(3x^2+4x)^5

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Example Question #1372 : Calculus

Differentiate:

$y=\sqrt{1+x^2}$ .

Possible Answers:

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

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

$y'=\frac{1}{2}x^2$

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

$y{}'={x}{\sqrt{1+x^2}}$

Correct answer:

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

Explanation:

It might be helpful to rewrite this equation first: $y=(1+x^2)^{\frac{1}{2}}$ . Then, apply the chain rule.

First, multiply the exponent by the coefficient in front of the expression (in this case, 1).

Then, subtract 1 from the exponent.

Lastly, take the derivative of the expression.

This comes out to

$y=\left(\frac{1}{2}\right)(1+x^2)^{\frac{-1}{2}}(2x)$ .

Then, multiply like terms and get rid of the negative exponent:

$y=\frac{x}{\sqrt{1+x^2}}$ .

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Calculus 1 : Other Differential Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #151 : How To Find Differential Functions

Example Question #334 : Differential Functions

Example Question #151 : How To Find Differential Functions

Example Question #151 : How To Find Differential Functions

Example Question #151 : Other Differential Functions

Example Question #341 : Differential Functions

Example Question #342 : Differential Functions

Example Question #152 : How To Find Differential Functions

Example Question #341 : Functions

Example Question #1372 : Calculus

All Calculus 1 Resources

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