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 $\frac{1}{1+x^2}.$