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

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Answer 1 · 2015-12-19T16:24:52

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

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:

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

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