# How do you find the derivative of f(x)=(x+3)/(x-3)?

Using the quotient rule we have that

f'(x)=((x+3)'*(x-3)-(x-3)'*(x+3))/(x-3)^2=> f'(x)=((x-3)-(x+3))/(x-3)^2=> f'(x)=-6/(x-3)^2

Feb 2, 2016

$- \frac{6}{x - 3} ^ 2$

#### Explanation:

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

for a rational function $f \left(x\right) = g \frac{x}{h \left(x\right)}$

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

applying this to the above function gives :

$\frac{d}{\mathrm{dx}} \left(\frac{x + 3}{x - 3}\right) = \frac{\left(x - 3\right) \frac{d}{\mathrm{dx}} \left(x + 3\right) - \left(x + 3\right) \frac{d}{\mathrm{dx}} \left(x - 3\right)}{x - 3} ^ 2$

$= \frac{\left(x - 3\right) .1 - \left(x + 3\right) .1}{x - 3} ^ 2 = \frac{x - 3 - x - 3}{x - 3} ^ 2$

$= - \frac{6}{x - 3} ^ 2$