# How do you differentiate f(x)=(x+3)^2/(x-1) using the quotient rule?

Jul 22, 2016

You have to first differentiate the numerator using the chain rule.

#### Explanation:

Let $y = {u}^{2}$ and $u = x + 3$. Then $y ' = 2 u$ and #u' = 1. The derivative of the numerator is :

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 u \times 1 = 2 \left(x + 3\right) = 2 x + 6$

Now, we can use the quotient rule to differentiate the entire function.

$f ' \left(x\right) = \frac{\left(2 x + 6\right) \left(x - 1\right) - 1 {\left(x + 3\right)}^{2}}{x - 1} ^ 2$

$f ' \left(x\right) = \frac{2 {x}^{2} + 6 x - 2 x - 6 - 1 \left({x}^{2} + 6 x + 9\right)}{{x}^{2} - 2 x + 1}$

$f ' \left(x\right) = \frac{{x}^{2} - 2 x - 15}{{x}^{2} - 2 x + 1}$

$f ' \left(x\right) = \frac{\left(x - 5\right) \left(x + 3\right)}{x - 1} ^ 2$

Hopefully this helps!