<s>[INST] Find the derivative of $y=\ln^3(3-x+x^2)$ Find the derivative of $$y=\ln^3(3-x+x^2)$$ We have to use the chain rule, but I am not sure how to think about it when we have more complex functions as in this problem. We can rewrite the function as $$y=\left[\ln(3-x+x^2)\right]^3,$$ so the outermost function is $g(x)=x^3$, right? And its derivative is $g'(x)=3x^2$. Then we have $t(x)=\ln(3-x+x^2)$ and $t'(x)=\dfrac{1}{3-x+x^2}(3-x+x^2)^{'}$ and finally $v(x)=(3-x+x^2)$ with $v'(x)=2x-1$. How do we use that in order to find $y'$? I would be grateful if you explained it in detail.
 [/INST] Another possible approach:
\begin{align*}
y = \ln^{3}(3 - x + x^{2}) & \Longleftrightarrow y^{1/3} = \ln(3 - x + x^{2})\\\\
& \Longleftrightarrow \frac{y'}{3y^{2/3}} = \frac{2x - 1}{3 - x + x^{2}}\\\\
& \Longleftrightarrow y' = \frac{3(2x - 1)\ln^{2}(3 - x + x^{2})}{3 - x + x^{2}}
\end{align*}
Hopefully this helps!
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