Solution
= 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> is
$$
\boxed{\frac{dy}{dx}=\frac1{1+x^2},\qquad x\in\mathbb R.}
$$
The <chain rule> gives the same result directly, using $1-x^2/(1+x^2)=1/(1+x^2)$ with a positive square root.