Find the derivative of $y=\sqrt{\ln\left(4x-x^2\right)}$ Find the derivative of $$y=\sqrt{\ln{\left(4x-x^2\right)}}$$
So we can rewrite the function as $$y=\left[\ln\left(4x-x^2\right)\right]^\frac12$$ Let's try to break it down a bit. So let's set $$a(x)=x^\frac12$$ and $$b(x)=\ln(4x-x^2)$$ then $$y=a(b(x))$$ The chain rule then tells us that $$y'=a'(b(x))b'(x)$$ Now $a'$ we can easily find, as it is just $$a'(x)=\dfrac{1}{2\sqrt{x}},$$ 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 $\beta(x)=4x-x^2$. These two functions we know how to differentiate! Indeed $\alpha'(x)=\dfrac{1}{x}$ and $\beta'(x)=4-2x$. Furthermore, the chain rule also tells us now that, as $b(x)=\alpha(\beta(x))$, $b'(x)=\alpha'(\beta(x))\beta'(x)$. Putting this all together we get that $$y'(x)=a'(b(x))\alpha'(\beta(x))\beta'(x)\\=\dfrac{1}{2\sqrt{\ln(4x-x^2)}}\cdot\dfrac{1}{4x-x^2}\cdot(4-2x)=\dfrac{1}{2x\sqrt{\ln(4x-x^2)}}.$$ The answer seems to differ...

$y'(x) =\frac{(\ln{{(4x-x^2)}})'}{2\times\sqrt{\ln{{4(x-x^2)}}}}=\frac{\frac{(4x-x^2)'}{4x-x^2}}{2\times\sqrt{\ln{{(4x-x^2)}}}}=\frac{\frac{4-2x}{x(4-x)}}{2\times\sqrt{\ln{{(4x-x^2)}}}}=\frac{\frac{x-2}{x(x-4)}}{\sqrt{\ln{{(4x-x^2)}}}}$.
Now,
write $\frac{x-2}{x(x-4)}$  such that $\frac A{x}+\frac B{x-2}=\frac{A(x-4)+Bx}{x(x-4)}=\frac{(A+B)x-4A}{x(x-4)}$.
This way,
$A+B=1$ ans $-4A=-2$, i.e. $A=1/2$ and $B=1/2$.
So,
$y'(x)=\frac{\frac{x-2}{x(x-4)}}{\sqrt{\ln{{(4x-x^2)}}}}=\frac{\frac 1{2x}+\frac 1{2(x-4)}}{\sqrt{\ln{{(4x-x^2)}}}}=\frac{1}{2x\sqrt{\ln{{(4x-x^2)}}}}+\frac{1}{2(x-4)\sqrt{\ln{{(4x-x^2)}}}}$.
