<s>[INST] Find the derivative of $f=\arcsin\left(\frac{2x}{1+x^2}\right)$ I'm trying to find the derivative of $f=\arcsin\left(\frac{2x}{1+x^2}\right)$. I think I'm mistaken and perhaps using the chain rule incorrectly. 
Let $g(x) = \frac{2x}{1+x^2}$ and let $h(x) = \arcsin x$
According to the chain rule - 
$$f'(x) = \frac{1}{\sqrt{1-\frac{2x}{1+x^2}}}⋅((2⋅(1+x^2 )-2x⋅2x)/(1+x^2 )^2 ) = \cdots \frac{-2(x^4-1)}{x-1}$$
Is this a correct usage of the chain rule?
 [/INST] If $\arctan x=y,-\dfrac\pi2\le y\le\dfrac\pi2, x=\tan y$
$\implies\dfrac{2x}{1+x^2}=\sin2y$
$\implies\arcsin\dfrac{2x}{1+x^2}=\begin{cases} 2\arctan x &\mbox{if }-\dfrac\pi2\le2y\le\dfrac\pi2\iff-1\le x\le1 \\\pi-2\arctan x & \mbox{if } 2y>\dfrac\pi2\iff x>1 \\-\pi-2\arctan x & \mbox{if } 2y<-\dfrac\pi2\iff x<-1\end{cases}$
Now $\dfrac{d(\arctan x)}{dx}=?$
Actually, $\sqrt{(1+x^2)^2-4x^2}=\sqrt{(1-x^2)^2}=|1-x^2|= \begin{cases} 1-x^2 &\mbox{if }1-x^2\ge0\iff-1\le x\le1\\
x^2-1& \mbox{if }x^2>1\iff x>1\text{ or }x<-1\end{cases}$
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