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Precalculus : Introductory Calculus

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Example Questions

Example Question #101 : Introductory Calculus

Find the first derivative of the function

f(x)=9x^6 .

Possible Answers:

f'(x)=54x^5

f'(x)=9

f'(x)=6x^5

f'(x)=9x^5

Correct answer:

f'(x)=54x^5

Explanation:

For the function

f(x)=ax^n

The first derivative is

$f^{'}(x)=anx^{n-1}$

So for

f(x)=9x^6

the first derivative is

$f'(x)=9(6)x^{6-1}=54x^5$

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Example Question #1 : Find The First Derivative Of A Function

Find the first derivative of

y(x)=x^2+x+2

in relation to .

Possible Answers:

x^2+x+1

2x^2+1

2x+1

Correct answer:

2x+1

Explanation:

To find the derviative of this equation recall the power rule that states: Multiply the exponent in front of the constant and then subtract one from the exponent.

$(x^{n})dx=nx^{n-1}$

We can work individually with each term:

Derivative of

x^2 is,

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

For the next term:

Derivative of x^1 :

1-1=0

So answer is:

1x^0

Anything to a power of 0 is 1.

For the next term:

Derivative of :

Any derivative of a constant is .

So the first derivative of

x^2+x+1

2x+1

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Example Question #61 : Derivatives

What is the derivative of

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

with respect to

Possible Answers:

$\frac{4x^2+8}{(x^2 +2)^2}$

$\frac{-4x^2+8}{(x^2+2)^2}$

$\frac{4}{x^2+2}$

$\frac{x^2-2}{(x^2+2)^2}$

Correct answer:

$\frac{-4x^2+8}{(x^2+2)^2}$

Explanation:

Recall the quotient rule:

$\left(\frac{f(x)}{g(x)}\right)' = \frac{g(x)f(x)'-{g(x)}'f(x)}{g(x)^2}$

So for:

f(x) =4x

g(x)=x^2+2

Find their derivatives:

{f(x)}'=4

{g(x)}'=2x

Plug in to the formula:

$\left(\frac{f(x)}{g(x)}\right)'=\frac{(x^2+2)(4)-4x(2x)}{(x^2+2)^2} =\frac{4x^2+8-8x^2}{(x^2+2)^2}=\frac{-4x^2+8}{(x^2+2)^2}$

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Example Question #62 : Derivatives

Find the first derivative of

y(x)=9x

with respect to

Possible Answers:

4.5x^2

9x^2

Correct answer:

Explanation:

Apply Power Rule.

First Derivative of 9x^1 :

$9\cdot1=9$

1-1=0

Result is

9x^0

So the first derivative of is

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Example Question #11 : Find The First Derivative Of A Function

Find the derivative of the following function:

f(x)=(3x^3-x^2)^4

Possible Answers:

f'(x)=4(9x^2-2x)^3

f'(x)=4(3x^3-x^2)^3(9x^2-2x)

f'(x)=4(3x^3-x^2)^3(9x^2-2x)(18x-2)

f'(x)=4(3x^3-x^2)^3

f'(x)=4(9x^2-2x)^3(3x^3-x^2)

Correct answer:

f'(x)=4(3x^3-x^2)^3(9x^2-2x)

Explanation:

To find the derivative of this function, we will have to apply both the power rule and the chain rule. First we apply the power rule, bringing the exponent of the entire term to the front as a coefficient and subtracting 1 from the new exponent. Then we apply the chain rule, multiplying that entire term by the derivative of just the expression inside the parentheses:

f(x)=(3x^3-x^2)^4

$f'(x)=4(3x^3-x^2)^3\frac{d}{dx}(3x^3-x^2)=4(3x^3-x^2)^3(9x^2-2x)$

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Example Question #14 : Find The First Derivative Of A Function

Find the derivative of the following function:

$f(x)=\frac{5}{2}x^3-\frac{2}{7}x^2+\ln (x)+18$

Possible Answers:

$f'(x)=5x^2-4x^{\frac{3}{2}}+x$

$f'(x)=\frac{5}{3}x^2-\frac{1}{7}x+\ln (x)$

$f'(x)=\frac{15}{2}x^2-\frac{4}{7}x+\frac{1}{x}$

$f'(x)=\frac{15}{4}x^2-\frac{3}{7}x+\frac{1}{2}\ln(x)^{-\frac{1}{2}}$

$f'(x)=\frac{5}{2}x^2-\frac{2}{7}x+18$

Correct answer:

$f'(x)=\frac{15}{2}x^2-\frac{4}{7}x+\frac{1}{x}$

Explanation:

To take the derivative of this function, we can simply apply the power rule to most of the terms. For the first two, we apply the power rule, multiplying the coefficient out front by the exponent of the term, and then subtracting 1 from the new exponent. When we get to ln(x), we must remember that the derivative of this term is 1/x. Finally, we have 18, which is just a constant, and the derivative of a constant is always 0:

$f(x)=\frac{5}{2}x^3-\frac{2}{7}x^2+\ln (x)+18$

$f'(x)=\frac{15}{2}x^2-\frac{4}{7}x+\frac{1}{x}$

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Example Question #11 : Find The First Derivative Of A Function

Find the derivative of the following function:

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

For the first term we can simply apply the power rule, multiplying the coefficient by the exponent of the term, which gives us 1 as the new coefficient, and subtracting one from the new exponent. When we get to our exponential terms, we have to remember to apply the chain rule to each one. The derivative of e^x is just e^x , because when we apply the chain rule, we multiply the term by the derivative of its exponent, which for x is just 1. For our next term, the exponent is 2x, so its derivative is 2, which we multiply by the whole term to give us $2e^{2x}$ . Finally our last term has an exponent of x^3 , which has a derivative of 3x^2 , so we multiply this by the whole term to give us $3x^2e^{x^3}$ :

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

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

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Example Question #11 : Find The First Derivative Of A Function

Find the derivative of $f(x)=7x^{2}+2x$ .

Possible Answers:

f'(x)=9x+2

f'(x)=14x+2

$f'(x)=14x^{2}+2$

$f(x)=9x^{2}+x$

$f(x)=9x^{2}+2x$

Correct answer:

f'(x)=14x+2

Explanation:

For any function $f(x)=Ax^{n}$ , the first derivative $f'(x)=nAx^{n-1}$ .

Therefore, taking each term of f(x) :

$f(x)=7x^{2}+2x$

$f'(x)=2(7x^{2-1})+1(2x^{1-1})$

f'(x)=14x+2

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Example Question #66 : Derivatives

Find the first derivative of $y=x^{3}$ .

Possible Answers:

$\frac{dy}{dx}=3x$

$y=3x^{2}$

$\frac{dx}{dy}=3x$

$\frac{dy}{dx}=3x^{2}$

y=3x

Correct answer:

$\frac{dy}{dx}=3x^{2}$

Explanation:

By the Power Rule of derivatives, for any equation $y=x^{a}$ , the derivative $\frac{dy}{dx}=a\cdot x^{a-1}$ .

With our function $y=x^{3}$ , where a=3 , we can therefore conclude that:

$\frac{dy}{dx}=3\cdot x^{3-1}$

$\frac{dy}{dx}=3x^{2}$

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Example Question #731 : Pre Calculus

Find the derivative of $y=4x^{5}$ .

Possible Answers:

$\frac{dy}{dx}=9x$

$\frac{dy}{dx}=9x^{4}$

$y=20x^{4}$

$\frac{dy}{dx}=20x^{4}$

$\frac{dy}{dx}=20x$

Correct answer:

$\frac{dy}{dx}=20x^{4}$

Explanation:

By the Power Rule of derivatives, for any equation $y=x^{a}$ , the derivative $\frac{dy}{dx}=a\cdot x^{a-1}$ .

Given our function $y=4x^{5}$ , where a=5 , we can conclude that

$\frac{dy}{dx}=5(4\cdot x^{5-1})$

$\frac{dy}{dx}=20x^{4}$

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Precalculus : Introductory Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #101 : Introductory Calculus

Example Question #1 : Find The First Derivative Of A Function

Example Question #61 : Derivatives

Example Question #62 : Derivatives

Example Question #11 : Find The First Derivative Of A Function

Example Question #14 : Find The First Derivative Of A Function

Example Question #11 : Find The First Derivative Of A Function

Example Question #11 : Find The First Derivative Of A Function

Example Question #66 : Derivatives

Example Question #731 : Pre Calculus

All Precalculus Resources

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