# How do you use the quotient rule to differentiate y=1/(x-4)^2?

Mar 4, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2}{x - 4} ^ 3$

#### Explanation:

differentiate using the $\textcolor{b l u e}{\text{ Quotient rule }}$

If $f \left(x\right) = g \frac{x}{h \left(x\right)} \text{then } f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2$

g(x) = 1 $\Rightarrow g ' \left(x\right) = 0$

and $h \left(x\right) = {\left(x - 4\right)}^{2} \Rightarrow h ' \left(x\right) = 2 \left(x - 4\right) \frac{d}{\mathrm{dx}} \left(x - 4\right)$

Replace these results into f'(x) :

$\Rightarrow f ' \left(x\right) = \frac{{\left(x - 4\right)}^{2} . 0 - 1 . 2 \left(x - 4\right)}{{\left(x - 4\right)}^{2}} ^ 2$

$= \frac{- 2 \left(x - 4\right)}{x - 4} ^ 4 = - 2 \frac{\cancel{x - 4}}{\cancel{x - 4}} ^ 3$