Question

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

Answer 1

`f'(x)=3x^2ln(x^3-4)+(3x^5)/(x^3-4)`

Explanation:

The first step should be to simplify as such:

`ln(x^3-4)^x=xln(x^3-4)`

so,

`f(x)=x^3ln(x^3-4)`

Now, we can use product rule.

`f'(x)=ln(x^3-4)d/dx[x^3]+x^3d/dx[ln(x^3-4)]`

Find each derivative separately.

`d/dx[x^3]=3x^2`

According to the chain rule, `d/dx[ln(u)]=(u')/u`.

`d/dx[ln(x^3-4)]=(d/dx[x^3-4])/(x^3-4)=(3x^2)/(x^3-4)`

Plug these back in.

`f'(x)=3x^2ln(x^3-4)+(3x^5)/(x^3-4)`

If you wish, this can be simplified further:

`f'(x)=(3x^2((x^3-4)ln(x^3-4)+x^3))/(x^3-4)`