# How does the quotient rule differ from the product rule?

Mar 29, 2015

The simple answer is that the
Quotient Rule:
$\frac{d \left(g \frac{x}{h \left(x\right)}\right)}{\mathrm{dx}} = \frac{g ' \left(x\right) h \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2$
is used when you want to find the derivative of a function which can
easily be viewed as one function divided by another function.
For example
$f \left(x\right) = \frac{3 {x}^{2} + 7}{\sqrt{x}}$
can be viewed as
$f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$
with $g \left(x\right) = 3 {x}^{2} + 7$ and $h \left(x\right) = \sqrt{x}$

The Product Rule:
$\frac{d \left(g \left(x\right)\right) \cdot \left(h \left(x\right)\right)}{\mathrm{dx}} = g ' \left(x\right) \cdot h \left(x\right) + g \left(x\right) \cdot h ' \left(x\right)$
is used when you want to find the derivative of a function which can be easily viewed as one function multiplied by another function.
For example
$f \left(x\right) = \left(3 {x}^{2} + 7\right) \cdot \sqrt{x}$

The more complex answer is that they aren't really different rules:
Given a function
$f \left(x\right) = \frac{g \left(x\right)}{r \left(x\right)}$
if you replace $r \left(x\right)$ with 1/(h(x) or $h {\left(x\right)}^{- 1}$
and apply the Product Rule
(with some care)
you will end up with the Quotient Rule.
[if you decide to try this you will need to remember $\frac{d \left(h {\left(x\right)}^{- 1}\right)}{\mathrm{dx}} = \left(- 1\right) \left(h {\left(x\right)}^{- 2}\right) \cdot \left(\frac{d h \left(x\right)}{\mathrm{dx}}\right)$