
# 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

Example $$\PageIndex{1}$$

Find

$\dfrac{d}{dx } (2 - x^2)(x^4 - 5) \nonumber$

Solution:

Here

$f(x) = 2 - x^2 \nonumber$

and

$g(x) = x^4 - 5 \nonumber$

The product rule gives

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

### 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

$\dfrac{d}{dx} \dfrac{f(x)}{g(x)} = \dfrac{g(x)\, f'(x) - f(x) \, g'(x)}{g(x)^2} \label{quot}$

Example $$\PageIndex{2}$$:

Find $$y'$$ if

$y' =\dfrac{2x - 1}{x + 1} \nonumber$

Solution

Here

$f(x) = 2x - 1 \nonumber$

and

$g(x) = x + 1 \nonumber$

The quotient rule (Equation \ref{quot}) gives

\begin{align*} \dfrac{ (x + 1)(2) - (2x - 1)(1)}{ (x + 1)2} &= \dfrac{2x + 2 - 2x + 1}{(x + 1)2} \\[5pt] &= \dfrac{3}{ (x + 1)2} \end{align*}