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

Dec 19, 2015

$f ' \left(x\right) = 3 {x}^{2} \ln \left({x}^{3} - 4\right) + \frac{3 {x}^{5}}{{x}^{3} - 4}$

Explanation:

The first step should be to simplify as such:

$\ln {\left({x}^{3} - 4\right)}^{x} = x \ln \left({x}^{3} - 4\right)$

so,

$f \left(x\right) = {x}^{3} \ln \left({x}^{3} - 4\right)$

Now, we can use product rule.

$f ' \left(x\right) = \ln \left({x}^{3} - 4\right) \frac{d}{\mathrm{dx}} \left[{x}^{3}\right] + {x}^{3} \frac{d}{\mathrm{dx}} \left[\ln \left({x}^{3} - 4\right)\right]$

Find each derivative separately.

$\frac{d}{\mathrm{dx}} \left[{x}^{3}\right] = 3 {x}^{2}$

According to the chain rule, $\frac{d}{\mathrm{dx}} \left[\ln \left(u\right)\right] = \frac{u '}{u}$.

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

Plug these back in.

$f ' \left(x\right) = 3 {x}^{2} \ln \left({x}^{3} - 4\right) + \frac{3 {x}^{5}}{{x}^{3} - 4}$

If you wish, this can be simplified further:

$f ' \left(x\right) = \frac{3 {x}^{2} \left(\left({x}^{3} - 4\right) \ln \left({x}^{3} - 4\right) + {x}^{3}\right)}{{x}^{3} - 4}$