# How do you differentiate (x^3 - 9x)/(x^2-7x+12)?

Feb 17, 2015

First, let us simplify the expression

(x(x^2-9x))/((x-4)(x-3)

(x(x+3)(x-3))/((x-4)(x-3)

$\frac{x \left(x + 3\right)}{x - 4}$

$\frac{{x}^{2} + 3 x}{x - 4}$

Now differentiate using the quotient rule.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left[\left(x - 4\right) \left(2 x + 3\right)\right] - \left[\left({x}^{2} + 3 x \left(1\right)\right]\right)}{x - 4} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {x}^{2} - 5 x - 12 - {x}^{2} - 3 x}{x - 4} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{2} - 8 x - 12}{x - 4} ^ 2$