Answer
$y'=\frac{1}{2+2x^{2}}$
Work Step by Step
By the chain rule it follows:
$$y'=(x-\sqrt{1+x^{2}})'(\arctan)'(x-\sqrt{1+x^{2}})$$
$$y'=(x-\sqrt{1+x^{2}})'\frac{1}{1+(x-\sqrt{1+x^{2}})^{2}}$$
$$y'=(1-\frac{2x}{2\sqrt{1+x^{2}}})\cdot \frac{1}{1+(x-\sqrt{1+x^{2}})^{2}}$$
$$y'=(1-\frac{x}{\sqrt{1+x^{2}}})\cdot \frac{1}{1+(x-\sqrt{1+x^{2}})^{2}}$$
After simplification it follows:
$$y'=\frac{1}{2+2x^{2}}$$
