# What is the derivative of f(x)=(x^2-4)ln(x^3/3-4x)?

Dec 30, 2015

We'll need the product rule for $f \left(x\right)$ and the chain rule for the second term $\ln \left({x}^{3} / 3 - 4 x\right)$

#### Explanation:

• Product rule: be $y = f \left(x\right) g \left(x\right)$, then $\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$
• Chain rule: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

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

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = 2 x \ln \left({x}^{3} / 3 - 4 x\right) + {\left({x}^{2} - 4\right)}^{2} / \left(\frac{{x}^{3} - 12 x}{3}\right)$

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = 2 x \ln \left({x}^{3} / 3 - 4 x\right) + \frac{3 {\left({x}^{2} - 4\right)}^{2}}{x \left({x}^{2} - 12\right)}$