# How do you differentiate f(x)=(x^2-2x)/(x+3)^2 using the quotient rule?

$f ' \left(x\right) = \frac{\left(2 x - 2\right) {\left(x + 3\right)}^{2} - 2 \left({x}^{2} - 2 x\right) \left(x + 3\right)}{x + 3} ^ 4 = \frac{\mathrm{df}}{\mathrm{dx}}$
You know that the derivative of the quotient of two functions $u$ and $v$is given by the formula $\frac{u ' v - u v '}{v} ^ 2$.
Here, $u \left(x\right) = {x}^{2} - 2 x$ and $v \left(x\right) = {\left(x + 3\right)}^{2}$ so $u ' \left(x\right) = 2 x - 2$ and $v ' \left(x\right) = 2 \left(x + 3\right)$ by the power rule. Hence the result.