Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

The Product and Quotient Rules

Theorem: The Product Rule

Let f and g be differentiable functions. Then

\[ \left[f(x) \, g(x)\right] ' = f(x)\, g '(x) + f '(x) \,g(x) \]

Proof

We have 

prodqu1.gif

Example \(\PageIndex{1}\):

Find

\[dfrac{d}{dx } (2 - x^2)(x^4 - 5)\]

Solution:

Here

\[ f(x) = 2 - x^2 \] and

\[  g(x) = x^4 - 5\]

The product rule gives

\[dfrac{d}{dx } (2 - x^2)(x^4 - 5) = (2 - x^2)(4x^3) + (-2x)(x^4 - 5)\]

The Quotient Rule

Remember the poem

"lo d hi minus hi d lo square the bottom and away you go"

This poem is the mnemonic for the taking the derivative of a quotient.

Theorem: The Quotient Rule

Let f and g be differentiable functions. Then

\[ \dfrac{d}{dx} \dfrac{f(x)}{g(x)} = \dfrac{g(x)\, f'(x) - f(x) \, g'(x)}{g(x)^2} \]

Example \(\PageIndex{2}\):

Find y' if \[ y' =\dfrac{2x - 1}{x + 1}\]

Solution: Here

\[ f(x) = 2x - 1\]

and

\[ g(x) = x + 1\] The quotient rule gives

\[ \dfrac{ (x + 1)(2) - (2x - 1)(1)}{ (x + 1)2}\]  

\[= \dfrac{2x + 2 - 2x + 1}{(x + 1)2}\]

\[= \dfrac{3}{ (x + 1)2}\]  

Contributors