# How do you find the derivative of y(x)= (9x)/(x-3)^2?

Aug 7, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{9 x + 27}{x - 3} ^ 3$

#### Explanation:

$\text{one way is to differentiate using the quotient rule}$

$\text{given "y=(g(x))/(h(x))" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2 \leftarrow \text{ quotient rule}$

$g \left(x\right) = 9 x \Rightarrow g ' \left(x\right) = 9$

$h \left(x\right) = {\left(x - 3\right)}^{2} \Rightarrow h ' \left(x\right) = 2 \left(x - 3\right)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{\left(x - 3\right)}^{2} .9 - 9 x .2 \left(x - 3\right)}{x - 3} ^ 4$

$\textcolor{w h i t e}{\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}} = \frac{\left(x - 3\right) \left(9 x - 27 - 18 x\right)}{x - 3} ^ 4$

$\textcolor{w h i t e}{\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}} = - \frac{9 x + 27}{x - 3} ^ 3$