Answer
$$y'=\frac{x}{\sqrt{1+x^2}}\cos\sqrt{1+x^2}$$
Work Step by Step
$$y'=(\sin\sqrt{1+x^2})'=\cos\sqrt{1+x^2}(\sqrt{1+x^2})'=
\cos\sqrt{1+x^2}\frac{1}{2\sqrt{1+x^2}}(1+x^2)'=
\cos\sqrt{1+x^2}\frac{1}{2\sqrt{1+x^2}}\cdot2x=\frac{x}{\sqrt{1+x^2}}\cos\sqrt{1+x^2}$$
