<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] Since it's easy to graph the function with some software (also online) it should be clear that the derivative cannot be everywhere positive and the graph also suggests the function is not differentiable at $-1$ and $1$.

Set $g(x)=\dfrac{2x}{1+x^2}$; then
$$
f(x)=\arcsin\dfrac{2x}{1+x^2}=\arcsin(g(x))
$$
and, by the chain rule,
$$
f'(x)=\frac{1}{\sqrt{1-(g(x))^2}}g'(x)
$$
Now divide et impera:


*

*$\displaystyle 1-(g(x))^2=\frac{(1+x^2)^2-4x^2}{(1+x^2)^2}
    =\frac{(1-x^2)^2}{(1+x^2)^2}$

*$\displaystyle \frac{1}{\sqrt{1-(g(x))^2}}=\frac{1+x^2}{|1-x^2|}$

*$\displaystyle g'(x)=2\frac{1+x^2-x\cdot2x}{(1+x^2)^2}$
Therefore
$$
f'(x)=\frac{1+x^2}{|1-x^2|}\frac{2(1-x^2)}{(1+x^2)^2}=
\frac{2}{1+x^2}\frac{1-x^2}{|1-x^2|}
$$
The derivative does exist at $-1$ and $1$, as it's easy to check with the limits from the left and from the right.

By the way, this shows that
$$
f(x)=\begin{cases}
c_--2\arctan x & \text{if $x<-1$} \\
c_0+2\arctan x & \text{if $-1\le x\le 1$}\\
c_+-2\arctan x & \text{if $x>1$}
\end{cases}
$$
Since
$$
\lim_{x\to-\infty}f(x)=0,\qquad
f(0)=0,\qquad
\lim_{x\to\infty}f(x)=0
$$
we can conclude that
$$
c_-=-\pi,\qquad c_0=0,\qquad c_+=\pi
$$
</s>