# What is the derivative of (x^2 - 4x)/(x-2)^2?

Jun 18, 2016

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{8}{x - 2} ^ 3$

#### Explanation:

To find derivative of $f \left(x\right) = \frac{{x}^{2} - 4 x}{x - 2} ^ 2$, we can use quotient rule.

According to quotient rule if $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{h \left(x\right) \cdot g ' \left(x\right) - g \left(x\right) \cdot h ' \left(x\right)}{h \left(x\right)} ^ 2$

Hence as $f \left(x\right) = \frac{{x}^{2} - 4 x}{x - 2} ^ 2$

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{{\left(x - 2\right)}^{2} \cdot \left(2 x - 4\right) - \left({x}^{2} - 4 x\right) \cdot 2 \left(x - 2\right)}{x - 2} ^ 4$

Now we can divide each term by $\left(x - 2\right)$ (as we already have a hole, a discontinuity, at $x = 2$ and at other places $x - 2 \ne 0$) and this results in

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

= $\frac{2 {x}^{2} - 8 x + 8 - 2 {x}^{2} + 8 x}{x - 2} ^ 3$

= $\frac{8}{x - 2} ^ 3$