Precalculus : Find the First Derivative of a Function

Study concepts, example questions & explanations for Precalculus

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Example Questions

Example Question #132 : Introductory Calculus

Find the first derivative of the following function:

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

Possible Answers:

\displaystyle f'(x)=5+14x

\displaystyle f'(x)=-14x+5

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

\displaystyle f'(x)=5x+14

Correct answer:

\displaystyle f'(x)=-14x+5

Explanation:

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

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

Thus,

\displaystyle 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:

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

Thus, our answer is:

\displaystyle f'(x)=5-7x

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

Find the first derivative of the following equation:

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

Possible Answers:

\displaystyle 2x+7

\displaystyle 2

\displaystyle 2x+77

\displaystyle 0

Correct answer:

\displaystyle 2x+77

Explanation:

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

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

Thus,

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

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

Find the first derivative of the following function:

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

Possible Answers:

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

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

\displaystyle 0

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

Correct answer:

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

Explanation:

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

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

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

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

Thus,

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

Distribute the 2x and 2.

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

Combine like terms to simplify.

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

Example Question #758 : Pre Calculus

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

Possible Answers:

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

\displaystyle f(x) = ln(x)

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

Correct answer:

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

Explanation:

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

Example Question #759 : Pre Calculus

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

Possible Answers:

None of the other answers.

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

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

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

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

Correct answer:

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

Explanation:

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

Simplify first:

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

Use the power rule on each term:

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

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

Note that constants become zero.

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