Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 1 / a / Solution

For real $x$, the angle $\arctan x$ lies in $(-\pi /2,\pi /2)$ and has sine $x/\sqrt{1+x^2}$. Thus the principal arcsine in the expression equals the arctangent of $x$. Its derivative isThe chain rule gives the same result directly, using $1-x^2/(1+x^2)=1/(1+x^2)$ with a positive square root.