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.
 A: Let's try to break it down a bit. So let's set
$$f(x)=\ln^3(3-x+x^2).$$
As you noticed first, we have "some function" to the power of $3$, and so if we set
$$a(x)=x^3$$
and
$$b(x)=\ln(3-x+x^2),$$
then
$$f(x)=a(b(x)).$$
The chain rule then tells us that
$$f'(x)=a'(b(x))b'(x).$$
Now $a'$ we can easily find, as it is just
$$a'(x)=3x^2,$$
but how do we find $b'$? Well, let's do the same thing again, namely notice that it's a composition and break it down. So now let
$$\alpha(x)=\ln(x)$$
and
$$\beta(x)=3-x+x^2.$$
These two functions we know how to differeniate! Indeed
$$\alpha'(x)=\frac{1}{x}$$
and
$$\beta'(x)=-1+2x.$$
Furthermore, the chain rule also tells us now that, as
$$b(x)=\alpha(\beta(x)),$$
we have that
$$b'(x)=\alpha'(\beta(x))\beta'(x).$$
Putting this all together we get that
$$f'(x)=a'(b(x))\alpha'(\beta(x))\beta'(x)=3\ln^2(3-x+x^2)\cdot\frac{1}{3-x+x^2}\cdot(-1+2x)=\frac{3(2x-1)\ln^2(3-x+x^2)}{3-x+x^2}.$$
A: 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!
A: You are almost done indeed, by chain rule, we have that
$$y=\left(\ln(g(x)\right)^3\implies y'=3\left(\ln(g(x)\right)^2\cdot (\ln(g(x))'\cdot g'(x)=3\left(\ln(g(x)\right)^2\cdot \frac 1 {g(x)}\cdot g'(x)$$
that is in your case
$$y'=3\left(\ln(3-x+x^2)\right)^2\cdot \frac{1}{3-x+x^2} \cdot (-1+2x)$$
