# How do you differentiate y=(x+3)^2/(x-1)?

May 13, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(x - 5\right) \left(x + 3\right)}{x - 1} ^ 2$

#### Explanation:

$y = {\left(x + 3\right)}^{2} / \left(x - 1\right)$

The quotient rule states that if $y = f \frac{x}{g} \left(x\right)$

Then $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{g \left(x\right) \cdot f ' \left(x\right) - f \left(x\right) \cdot g ' \left(x\right)}{g \left(x\right)} ^ 2$

Using the quotient rule in this example:
$\frac{\mathrm{dy}}{\mathrm{dx}}$= $\frac{\left(x - 1\right) \cdot 2 \left(x + 3\right) \cdot 1 - {\left(x + 3\right)}^{2} \cdot 1}{x - 1} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \left({x}^{2} + 2 x - 3\right) - \left({x}^{2} + 6 x + 9\right)}{x - 1} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{2} - 2 x - 15}{x - 1} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(x - 5\right) \left(x + 3\right)}{x - 1} ^ 2$