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?
 A: Your mistake here.
$$\sin^{-1} \frac{2x}{1+x^2} = \frac1{\sqrt{1-\left(\frac{2x}{1+x^2}\right)^2}}.$$
Its not $\sin^{-1} \dfrac{2x}{1+x^2} = \dfrac1{\sqrt{1-\frac{2x}{1+x^2}}}$.
After this your quotient rule is fine.
A: With the chain rule:
$$f'(x)=\frac{1}{\sqrt{1-\left(\frac{2x}{1+x^2}\right)^2}}\left(\frac{2(1+x^2)-2x(2x)}{(1+x^2)^2}\right)=\frac{1+x^2}{\sqrt{(1+x^2)^2-4x^2}}\left(\frac{2-2x^2}{(1+x^2)^2}\right)$$
$$=\frac{2-2x^2}{1+x^2}\frac{1}{|1-x^2|}=\frac{2}{1+x^2}\frac{1-x^2}{|1-x^2|}$$
A: 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
$$
c_-=-\pi,\qquad c_0=0,\qquad c_+=\pi
$$
A: 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, $\sqrt{(1+x^2)^2-4x^2}=\sqrt{(1-x^2)^2}=|1-x^2|= \begin{cases} 1-x^2 &\mbox{if }1-x^2\ge0\iff-1\le x\le1\\
x^2-1& \mbox{if }x^2>1\iff x>1\text{ or }x<-1\end{cases}$
A: \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.
$$
\frac d {dx} \arcsin\left( \frac{2x}{1+x^2}\right) = \begin{cases} \dfrac 2 {1+x^2} & \text{if } -1<x<1, \\[8pt] \dfrac{-2}{1+x^2} & \text{if } x>1 \text{ or } x<-1. \end{cases}
$$
