# How do you differentiate f(x) = 1/(x^3-4x) using the quotient rule?

Dec 24, 2015

Quotient rule states that for a function $y = f \frac{x}{g} \left(x\right)$, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{g} {\left(x\right)}^{2}$.

To be honest, I'd better use chain rule, here, but let's do as you asked.

#### Explanation:

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{\left(0\right) \left({x}^{3} - 4 x\right) - 1 \left(3 {x}^{2} - 4\right)}{{x}^{3} - 4 x} ^ 2$

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{\cancel{\left(0\right) \left({x}^{3} - 4 x\right)} - 1 \left(3 {x}^{2} - 4\right)}{{x}^{3} - 4 x} ^ 2$

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{- 1 \left(3 {x}^{2} - 4\right)}{{x}^{3} - 4 x} ^ 2 = \frac{4 - 3 {x}^{2}}{{x}^{3} - 4 x} ^ 2$