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Calculus 1 : How to find differential functions

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Example Questions

Example Question #141 : How To Find Differential Functions

Differentiate the function:

f(z)=az^-^7

Possible Answers:

f'(z)=-7z^-^8

f'(z)=-7az^-^6

f'(z)=-7az^-^8

f'(z)=7az^6

Correct answer:

f'(z)=-7az^-^8

Explanation:

To differentiate this problem we will need to use the power rule.

The power rule is, nx^n^-^1 where n is the exponent.

Thus our derivative is,

$f'(z)=\frac{d}{dz}(az^-^7)=-7\cdot az^-^7^-^1=-7az^-^8$ .

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Example Question #142 : How To Find Differential Functions

Differentiate the function:

$f(x)=-12x^\frac{-3}{4}$

Possible Answers:

$f'(x)=9x^\frac{1}{4}$

$f'(x)=9x^\frac{-7}{4}$

$f'(x)=12x^\frac{-3}{4}$

$f'(x)=-9x^\frac{1}{4}$

Correct answer:

$f'(x)=9x^\frac{-7}{4}$

Explanation:

To find the derivative of this function we will need to use the power rule.

The power rule is nx^n^-^1 where n is the exponent.

Therefore, our derivative is

$f'(x)=\left(-\frac{3}{4}\right)\cdot (-12)x^\frac{-3}{4}^-^{1}=9x^{\frac{-7}{4}}$ .

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Example Question #143 : How To Find Differential Functions

Differentiate the function:

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

Roots of numbers are fractional exponents so express $\sqrt{x}$ as $x^{\frac{1}{2}}$ and differentiate each term with respect to x using the power rule:

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$

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

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Example Question #143 : How To Find Differential Functions

Differentiate the function:

$f(x)=1000x^{1000}$

Possible Answers:

$f'(x)=1000000x^{999}$

$f'(x)=\infty$

$f'(x)=1000x^{999}$

f'(x)=1000000x

Correct answer:

$f'(x)=1000000x^{999}$

Explanation:

To find the derivative of this function we will need to use the power rule.

The power rule is nx^n^-^1 where n is the exponent.

Therefore, our derivative is

$f'(x)=1000\cdot 1000x^{1000-1}=1000000x^{999}$

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Example Question #145 : How To Find Differential Functions

A farmer wants to enclose a 360,000 square feet of land with as little fence as possible. Find the smalest perimeter of a fence possible to enclose 360,000 square feet of land.

Possible Answers:

None of the above.

2800feet

2400 feet

2600 feet

1200 feet

Correct answer:

2400 feet

Explanation:

To solve this question we must first realize that the area of a rectangle is

A=l*w

and the perimeter of a rectangle is

P=2l+2w

To start this question off, we must first solve the length in terms of the width, or vice versa from the first equation.

$l=\frac{360,000}{w}$

Then plug this into the perimeter equation

$P=\frac{720,000}{w}+2w$

In order to find the perimeter at its minimum, we must take the derivative of the perimeter equation with respect to , set it equal to zero, then find the value of . To take the derivative of this equation, we must utilize the power rule, $\frac{d}{dx}x^n=nx^{n-1}$ .

$0=\frac{-720,000}{w^2}+2$

$w=\sqrt{360,000}=600$

Plugging back into the original equation to solve for

360,000=600*l

l=600

Therefore the smallest possible perimeter to enclose 360,000 square feet of land is

2(600)+2(600)=2400 feet

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Example Question #146 : How To Find Differential Functions

Find the derivative of $\frac{\sin(\tan(x))}{x^2}$ .

Possible Answers:

$\frac{\sec^2(x)\cos(\tan(x))}{2x}$

$\frac{x^2\cos(\tan(x)) - 2\sin(\tan(x))}{x^3}$

$\frac{x\sec^2(x))\cos(\tan(x)) - 2\sin(\tan(x))}{x}$

$\frac{2\sin(\tan(x)) - \sec^2(x)\cos(\tan(x))}{x^3}$

$\frac{x\sec^2(x))\cos(\tan(x)) - 2\sin(\tan(x))}{x^3}$

Correct answer:

$\frac{x\sec^2(x))\cos(\tan(x)) - 2\sin(\tan(x))}{x^3}$

Explanation:

To find the derivative of this function, the quotient rule can be applied.

$u(x) = 2\sin(\tan(x)) - \sec^2(x)\cos(\tan(x))$

v(x) = x^2

Applying the quotient rule,

$f'(x) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{v(x)^2}$

After simplification, we have

$\frac{x\sec^2(x))\cos(\tan(x)) - 2\sin(\tan(x))}{x^3}$

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Example Question #147 : How To Find Differential Functions

Differentiate the function:

f(x)=x^3(1-3x)

Possible Answers:

f'(x)=3x^2-12x^3

f'(x)=(1-3x)x^2

f'(x)=-6x^2

f'(x)=x^3-3x^4

Correct answer:

f'(x)=3x^2-12x^3

Explanation:

Take away the parentheses by multiplying the x^3 to each term inside the parentheses. This will leave a polynomial which will make the differentiation simplier.

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

$\\f'(x)=\frac{d}{dx}[x^3(1-3x)]\\ \\ =\frac{d}{dx}(x^3-3x^4)\\ \\=\frac{d}{dx}(x^3)-\frac{d}{dx}(3x^4)\\ \\=3\cdot x^{3-1}-4\cdot 3x^{4-1}\\ \\=3x^2-12x^3$

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Example Question #148 : How To Find Differential Functions

Differentiate the function:

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

Possible Answers:

f'(x)=2

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

f'(x)=2x-3

f'(x)=4x-1

Correct answer:

f'(x)=4x-1

Explanation:

Take away the parentheses by multiplying the two binomials together so that you are left with a polynomial.

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

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

$\\f'(x)=\frac{d}{dx}(2x^2-x-6)\\ \\=\frac{d}{dx}(2x^2)-\frac{d}{dx}(x)-\frac{d}{dx}(6)\\ \\=2\cdot 2x^{2-1}-1\cdot x^{^{1-1}}\\ \\=4x-1$

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Example Question #144 : How To Find Differential Functions

Differentiate the function:

$f(x)=-\frac{12}{x^3}$

Possible Answers:

f'(x)=0

$f'(x)=-12x^{-3}$

f'(x)=36x^-^4

f'(x)=-36x^2

Correct answer:

f'(x)=36x^-^4

Explanation:

Bring the variable to the top to obtain one term and apply the power rule $nx^{n-1}$ where n is the exponent:

$f(x)=-\frac{12}{x^3}=-12x^-^3$

$\\f'(x)=\frac{d}{dx}(-12x^{-3})\\ \\=(-3)(-12x^{-3-1})\\ \\=36x^-^4$

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Example Question #150 : How To Find Differential Functions

Differentiate the function:

$f(a)=a^{\frac{5}{3}}-a^{\frac{2}{3}}$

Possible Answers:

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

$f'(a)=a^{\frac{7}{3}}$

$f'(a)=\sqrt[3]{x^5}-\sqrt[3]{x^2}$

f'(a)=a

Correct answer:

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

Explanation:

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

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

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Calculus 1 : How to find differential functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #141 : How To Find Differential Functions

Example Question #142 : How To Find Differential Functions

Example Question #143 : How To Find Differential Functions

Example Question #143 : How To Find Differential Functions

Example Question #145 : How To Find Differential Functions

Example Question #146 : How To Find Differential Functions

Example Question #147 : How To Find Differential Functions

Example Question #148 : How To Find Differential Functions

Example Question #144 : How To Find Differential Functions

Example Question #150 : How To Find Differential Functions

All Calculus 1 Resources

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