<s>[INST] 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 $h'(x)$?
 [/INST] Something to notice is that $$x^4 - 6x^2 + 9 = (x^2 - 3)^2$$ and consequently, $$f(x) = 2 (x^3 - 4x + 6) \log (x^2 - 3).$$  This makes differentiation slightly easier:  $$f'(x) = 2\left( (3x^2 - 4)\log(x^2 - 3) + \frac{x^3-4x+6}{x^2-3} \cdot 2x \right).$$
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