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?

\begin{align}
\frac d {dx} \arcsin\left( \frac{2x}{1+x^2} \right) & = \frac 1 {\sqrt{1-\left( \frac{2x}{1+x^2} \right)^2}} \cdot \frac d {dx} \frac{2x}{1+x^2} & & \text{by the chain rule} \\[10pt]
& = \frac 1 {\sqrt{1-\left( \frac{2x}{1+x^2} \right)^2}} \cdot \frac{2(1-x^2)}{(1+x^2)^2} & & \text{by the quotient rule} \\[10pt]
& = \frac 1 {\sqrt{\frac{(1+x^2)^2 - (2x)^2}{(1+x^2)^2} }} \cdot \frac{2(1-x^2)}{(1+x^2)^2} \\[10pt]
& = \frac{1+x^2}{\sqrt{(1+x^2)^2 - (2x)^2}} \cdot \frac{2(1-x^2)}{(1+x^2)^2} \\[10pt]
\end{align}
Now notice that $$ (1+x^2)^2 - (2x)^2 = (1+2x^2+x^4) - 4x^2 = 1-2x^2 + x^4 = (1-x^2)^2 $$
so we get
\begin{align}
\sqrt{(1+x^2)^2 - (2x)^2} = |1-x^2|.
\end{align}
If $-1\le x\le1$ then this is $1-x^2$ and so by cancellations from the numerator and denominator the derivative simplifies to
$$
\frac 2 {1+x^2} \qquad \left( = 2 \frac d {dx} \arctan x \right).
$$
However, since arcsine is not differentiable at $\pm1,$ the chain rule does not tell us that the composite function is differentiable at points where $2x/(1+x^2)\in\{\pm1\}.$ Those two points must therefore be examined separately.
If $x>1$ or $x<-1$ then $|1-x^2| = -(1-x^2)$ and the answer we get will be multiplied by $-1$.
Notice that $2x/(1+x^2)=\pm1$ when $x=\pm1$ and has absolute extreme values at those points. Thus $2x/(1+x^2)$ is never outside the domain of the arcsine function.
So the derivative is $\dfrac{-2}{1+x^2}$ when $x>1$ or $x<-1$.
When $x=\pm1$ then $2x/(1+x^2)=\pm1$ and one can find the two one-sided derivatives at those points. Lo and behold, they are not equal, so the function is not differentiable at those two points.
$$
\frac d {dx} \arcsin\left( \frac{2x}{1+x^2}\right) = \begin{cases} \dfrac 2 {1+x^2} & \text{if } -1<x<1, \\[8pt] \dfrac{-2}{1+x^2} & \text{if } x>1 \text{ or } x<-1. \end{cases}
$$
