# How do you differentiate f(x) = (x^2-4x)/(x+1) using the quotient rule?

$f ' \left(x\right) = \frac{\left(2 x - 4\right) \left(x + 1\right) - {x}^{2} + 4 x}{x + 1} ^ 2$
Let $f \left(x\right) = \frac{u \left(x\right)}{v \left(x\right)}$ where $u \left(x\right) = {x}^{2} - 4 x$ and $v \left(x\right) = x + 1$.
By the quotient rule, $f ' \left(x\right) = \frac{u ' \left(x\right) v \left(x\right) - u \left(x\right) v ' \left(x\right)}{v \left(x\right)} ^ 2$. Here, $u ' \left(x\right) = 2 x - 4$ and $v ' \left(x\right) = 1$.
So $f ' \left(x\right) = \frac{\left(2 x - 4\right) \left(x + 1\right) - {x}^{2} + 4 x}{x + 1} ^ 2$ by direct use of the quotient rule.