How to find the derivative of $f(x)=(x^3-4x+6)\ln(x^4-6x^2+9)$? Find the derivative of the following: $$f(x)=(x^3-4x+6)\ln(x^4-6x^2+9)$$ Would I use the chain rule and product rule?
So far I have:
$$\begin{align}g(x)=x^3-4x+6
\\g'(x)=2x^2-4\end{align}$$
would $h(x)$ be $\ln(x^4-6x^2+9)$?
If so, how would I find $h'(x)$?

given $f(x)=(x^3-4*x+6)\ln(x^4-6x^2+9)$ we find by the product and chaine rule
$f'(x)=(3x^2-4)\ln(x^4-6x^2+9)+(x^3-4x+6)\frac{4x^3-12x}{x^4-6x^2+9}$
simplifying this in a few minutes
simplifying this i got
$f'(x)=\frac{4 x^4-16 x^2+3 x^4 \ln \left(x^4-6 x^2+9\right)-13 x^2 \ln \left(x^4-6
   x^2+9\right)+12 \ln \left(x^4-6 x^2+9\right)+24 x}{x^2-3}$
