How can I differentiate this? I was trying to differentiate this: $\frac{1}{2} \sin^{-1} \frac{2x}{1+x^2}$ but I am really stuck.I obtained an answer that does not match with the one given in the book.Your help is appreciated.[Edit: I used the formula $\frac{d}{dx} \sin^{-1}x=\frac{1}{\sqrt{1-x^2}}$ to get $\frac{1+x^2}{2(1-x^2)}$]
Thank you.
 A: The derivative of the arcsine function is
$$\frac{d}{du}\arcsin(u) = \frac{1}{\sqrt{1-u^2}}.$$
So, using the Chain Rule, you would have
$$\frac{d}{dx}\frac{1}{2}\arcsin\left(\frac{2x}{1+x^2}\right) = \frac{1}{2}\left(\frac{1}{\sqrt{1 - \left(\frac{2x}{1+x^2}\right)^2}}\right)\left(\frac{2x}{1+x^2}\right)'.$$
The derivative of $\frac{2x}{1+x^2}$ is a simple application of either the Quotient Rule or the Product, Power, and Chain Rules:\
$$\begin{align*}
\frac{d}{dx}\left(\frac{2x}{1+x^2}\right) &= \frac{(1+x^2)(2x)' - 2x(1+x^2)'}{(1+x^2)^2}\\
&= \frac{2(1+x^2) - 2x(2x)}{(1+x^2)^2}\\
&= \frac{2 + 2x^2 - 4x^2}{(1+x^2)^2}\\
&= \frac{2(1-x^2)}{(1+x^2)^2}.
\end{align*}$$
The expression obtained from the derivative of $\arcsin(u)$ can use a bit of simplification too:
$$\begin{align*}
\frac{1}{\sqrt{1 - \left(\frac{2x}{1+x^2}\right)^2}} &= \frac{1}{\sqrt{1 - \frac{4x^2}{(1+x^2)^2}}}\\
&=\frac{1}{\sqrt{\frac{(1+x^2)^2 -4x^2}{(1+x^2)^2}}}\\
&= \frac{1}{\frac{\sqrt{1+2x^2+x^4 - 4x^2}}{|1+x^2|}}\\
&= \frac{|1+x^2|}{\sqrt{1-2x^2+x^4}}\\
&= \frac{1+x^2}{\sqrt{(1-x^2)^2}}\\
&= \frac{1+x^2}{|1-x^2|}.
\end{align*}$$
(Remember that $\sqrt{a^2}=|a|$, not $a$; we can get rid of the absolute value bars around $1+x^2$ because it is always positive; the same is not true with $1-x^2$).
Putting it all together, we have:
$$\begin{align*}
\frac{d}{dx}\frac{1}{2}\arcsin\left(\frac{2x}{1+x^2}\right) &= \frac{1}{2}\left(\frac{1}{\sqrt{1 - \left(\frac{2x}{1+x^2}\right)^2}}\right)\left(\frac{2x}{1+x^2}\right)'\\
&= \frac{1}{2}\left(\frac{1+x^2}{|1-x^2|}\right)\left(\frac{2(1-x^2)}{(1+x^2)^2}\right)\\
&= \frac{1-x^2}{(1+x^2)|1-x^2|}.
\end{align*}$$
This can be rewritten with the "sign function",
$$\mathrm{sgn}(u) = \left\{\begin{array}{ll}
1 & \text{if }u\gt 0;\\
-1 &\text{if }u\lt 0.
\end{array}\right.$$
as
$$\frac{d}{dx}\frac{1}{2}\arcsin\left(\frac{2x}{1+x^2}\right) = \frac{\mathrm{sgn}(1-x^2)}{1+x^2}.$$
It's also possible your book was not careful with the square root of the square, so that the answer given is just
$$\frac{1}{1+x^2}.$$
A: $y=\sin(x)\Rightarrow\frac{\mathrm{d}y}{\mathrm{d}x}=\cos(x)=\sqrt{1-y^2}\therefore\frac{1}{\sqrt{1-y^2}}=\frac{\mathrm{d}x}{\mathrm{d}y}=\frac{\mathrm{d}}{\mathrm{d}y}\sin^{-1}(y)$
The rest is the chain-rule:
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\frac{1}{2}\sin^{-1}\left(\frac{2x}{1+x^2}\right)
&=\frac{1}{2}\frac{1}{\sqrt{1-\left(\frac{2x}{1+x^2}\right)^2}}\frac{\mathrm{d}}{\mathrm{d}x}\frac{2x}{1+x^2}\\
&=\frac{1}{2}\frac{1+x^2}{|1-x^2|}\frac{2(1+x^2)-2x(2x)}{(1+x^2)^2}\\
&=\frac{1}{2}\frac{1+x^2}{|1-x^2|}\frac{2(1-x^2)}{(1+x^2)^2}\\
&=\frac{\operatorname{sgn}(1-x^2)}{1+x^2}\tag{1}
\end{align}
$$
Let $x=\tan(\theta)$, then $\frac{2x}{1+x^2}=\sin(2\theta)$. When $|\theta|<\frac{\pi}{4}$, $|x|<1$, so $\operatorname{sgn}(1-x^2)=1$. Then, equation $(1)$ is the reciprocal of
$$
\frac{\mathrm{d}}{\mathrm{d}\theta}\tan(\theta)=\sec^2(\theta)
$$
When $|\theta|\in(\frac{\pi}{4},\frac{\pi}{2})$, $|x|>1$, so $\operatorname{sgn}(1-x^2)=-1$, but $\sin^{-1}(\sin(2\theta))=\pi\operatorname{sgn}(\theta)-2\theta$. Then, equation $(1)$ is the reciprocal of
$$
\frac{\mathrm{d}}{\mathrm{d}\theta}(-\tan(\theta))=-\sec^2(\theta)
$$
A: Another approach would first prove the trigonometric identity
$$
\frac12 \arcsin\frac{2x}{1+x^2} = \arctan x \text{ if } -1 \le x \le 1.
$$
Then you have to figure out separately what to do when $x>1$ or $x<-1$.  If I'm not mistaken, you get
$$
\frac12 \arcsin\frac{2x}{1+x^2} = \frac\pi2 -\arctan x \text{ if } x>1
$$
and
$$
\frac12 \arcsin\frac{2x}{1+x^2} = -\frac\pi2 -\arctan x \text{ if } x<-1.
$$
Then you just need to remember how to differentiate the arctangent function.
A: Shortest way around:
Put $x= tan \theta $ to simplify the expression inside the arcsin function to $sin 2\theta$ .
The value of the given function is then just $\theta$ which is equal to arctan $x$
 , which on differentiating gives $\frac {1}{1+x^2}$
