# 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...

• Your last equality is incorrect, $x=2$ is a root of $4-2x$ but not of $4x-x^2$ so the last simplification cannot occur. Commented Aug 27, 2022 at 22:26
• It looks like you simplified $\dfrac{1}{4x-x^2}\cdot(4-2x)$ to $1/x$. It actually equals $\frac{1}{x-4}+\frac{1}{x}$. Commented Aug 27, 2022 at 22:29
• Silly me, thank you! Commented Aug 27, 2022 at 22:32

$$y'(x) =\frac{(\ln{{(4x-x^2)}})'}{2\times\sqrt{\ln{{4(x-x^2)}}}}=\frac{\frac{(4x-x^2)'}{4x-x^2}}{2\times\sqrt{\ln{{(4x-x^2)}}}}=\frac{\frac{4-2x}{x(4-x)}}{2\times\sqrt{\ln{{(4x-x^2)}}}}=\frac{\frac{x-2}{x(x-4)}}{\sqrt{\ln{{(4x-x^2)}}}}$$.
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, $$y'(x)=\frac{\frac{x-2}{x(x-4)}}{\sqrt{\ln{{(4x-x^2)}}}}=\frac{\frac 1{2x}+\frac 1{2(x-4)}}{\sqrt{\ln{{(4x-x^2)}}}}=\frac{1}{2x\sqrt{\ln{{(4x-x^2)}}}}+\frac{1}{2(x-4)\sqrt{\ln{{(4x-x^2)}}}}$$.