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Precalculus : Derivatives

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Example Questions

Example Question #132 : Introductory Calculus

Find the first derivative of the following function:

f(x)=5x-7x^2-2

Possible Answers:

f'(x)=5+14x

f'(x)=-14x+5

f'(x)=5x-7x^2-2

f'(x)=5x+14

Correct answer:

f'(x)=-14x+5

Explanation:

To solve, simply differentiate using the power rule for differentiation as outline below.

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

Thus,

$f'(x)=5x^{1-1}-7x^{2-1}$

Notice that our constant term disappeared because the derivative of a constant is zero.

Simplifying the above equation, we get:

f'(x)=5x^0-7x^1=5-7x

Thus, our answer is:

f'(x)=5-7x

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Example Question #41 : Find The First Derivative Of A Function

Find the first derivative of the following equation:

f(x)=x^2+77x-1

Possible Answers:

2x+7

2x+77

Correct answer:

2x+77

Explanation:

To solve, simply use the power rule as outline below:

Power rule: $(x^n)'=nx^{n-1}$

Thus,

$f'(x)=2x^{2-1}+77x^{1-1}=2x+77$

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Example Question #134 : Introductory Calculus

Find the first derivative of the following function:

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

Possible Answers:

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

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

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

Correct answer:

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

Explanation:

To solve, you must use the product rule as outline below.

Product rule: (g(x)*h(x))'=g(x)h'(x)+g'(x)h(x)

$g(x)=(2x+1)\rightarrow g'(x)=2$

$h(x)=(x^2-3)\rightarrow h'(x)=2x$

Thus,

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

Distribute the 2x and 2.

f'(x)=4x^2+2x+2x^2-6

Combine like terms to simplify.

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

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Example Question #758 : Pre Calculus

Find the first derivative of the function $f(x) = \frac{1}{x}$ .

Possible Answers:

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

f(x) = ln(x)

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

Correct answer:

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

Explanation:

An equivalent form of writing $f(x) = \frac{1}{x}$ is $f(x) = x^{-1}$ . The derivative of an exponential power x^b is $bx^{b-1}$ , so the derivative of $f(x) = x^{-1}$ is $f'(x) = -x^{-2}$ , or $f'(x) = \frac{-1}{x^2}$ .

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Example Question #759 : Pre Calculus

Find the first derivative: 12x^4-3x^3+2x^2-12-6x-9x^3

Possible Answers:

None of the other answers.

12x^3-3x^2+4x-18

48x^3-36x^2+4x+18

48x^3-36x^2+4x-6

48x^3-36x^2+4x-18

Correct answer:

48x^3-36x^2+4x-6

Explanation:

12x^4-3x^3+2x^2-12-6x-9x^3

Simplify first:

12x^4-12x^3+2x^2-12-6x

Use the power rule on each term:

Power rule: $x^a=ax^{a-1}$

48x^3-36x^2+4x-6

Note that constants become zero.

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Example Question #1 : Find The Critical Numbers Of A Function

What are the critical values of the function (x^3-2x^2+5) ?

Possible Answers:

None of the other answers

$x=0;x=\frac{4}{3}$

x=2

x=0

x=0, x=-2

Correct answer:

$x=0;x=\frac{4}{3}$

Explanation:

A number is critical if it makes the derivative of the expression equal 0.

Therefore, we need to take the derivative of the expression and set it to 0. We can use the power rule for each term of the expression.

$f^{'} = 3x^2 - 4x + 0 = 0$

Next, we need to factor the expression:

x(3x-4)= 0

We can now set each term equal to 0 to find the critical numbers:

x = 0

3x - 4 = 0

Therfore, our critical numbers are,

x=0

$x=\frac{4}{3}$

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Example Question #2 : Find The Critical Numbers Of A Function

Find the critical value(s) of the function.

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

Possible Answers:

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

$x=0,-\frac{1}{3}$

$x=\frac{1}{3},-\frac{1}{3}$

$x=\frac{1}{2},-\frac{1}{4}$

$x=\frac{1}{2},-\frac{2}{3}$

Correct answer:

$x=0,-\frac{1}{3}$

Explanation:

To find the critical values of a function, we must set the derivate equal to 0.

First, we find the derivative of the function to be

$\frac{dy}{dx}=3(6)x^2+2(3)x$

=18x^2+6x

We can then factor out a 6x and set the expression equal to 0

6x(3x+1)=0

From here, we can easily determine that

6x=0

x=0

and

3x+1=0

3x=-1

$x=-\frac{1}{3}$ .

Therefore, the critical values of the function are at x=0 and $x=-\frac{1}{3}$ .

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Example Question #1 : Find The Critical Numbers Of A Function

Find the critical points of the following function:

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

Possible Answers:

x=-4,x=5

x=-3,x=5

x=-5,x=3

x=-15,x=4

x=-4,x=15

Correct answer:

x=-5,x=3

Explanation:

The critical points of a function are the points at which its slope is zero, so first we must take the derivative of the function so we have a function that describes its slope:

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

f'(x)=x^2+2x-15

Now that we have the derivative, which tells us the slope of f(x) at any point x, we can set it equal to 0 and solve for x to find the points at which the slope of the function is 0, which are our critical points:

$x^2+2x-15=0\rightarrow (x+5)(x-3)=0\rightarrow x=-5,x=3$

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Example Question #1 : Find The Critical Numbers Of A Function

Find the critical number(s) of the function

f(x)=x^2 .

Possible Answers:

0.5

DNE

Correct answer:

Explanation:

To find the critical numbers, find the values for x where the first derivative is 0 or undefined.

For the function

f(x)=ax^n

The first derivative is

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

So for

f(x)=x^2

the first derivative is

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

Setting that equal to zero

2x=0

We find

x=0

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Example Question #1 : Find The Critical Numbers Of A Function

Find the critical number(s) of the function

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

Possible Answers:

DNE

0, 2

Correct answer:

0, 2

Explanation:

To find the critical numbers, find the values for x where the first derivative is 0 or undefined.

For the function

f(x)=ax^n .

The first derivative is

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

So for

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

the first derivative is

$f'(x)=3x^{3-1}-3(2)x^{2-1}=3x^2-6x=3x(x-2)$ .

Settig the first derivative equal to zero

f'(x)=3x(x-2) = 0

yields the values

x=0,2

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Precalculus : Derivatives

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #132 : Introductory Calculus

Example Question #41 : Find The First Derivative Of A Function

Example Question #134 : Introductory Calculus

Example Question #758 : Pre Calculus

Example Question #759 : Pre Calculus

Example Question #1 : Find The Critical Numbers Of A Function

Example Question #2 : Find The Critical Numbers Of A Function

Example Question #1 : Find The Critical Numbers Of A Function

Example Question #1 : Find The Critical Numbers Of A Function

Example Question #1 : Find The Critical Numbers Of A Function

All Precalculus Resources

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