Answer
$y'=\frac{\sqrt{1+x^2}-x}{(1+(x-\sqrt{1+x^2})^2)(\sqrt{1+x^2})}=\frac{1}{2(1+x^2)}$
Work Step by Step
$y=tan^{-1}(x-\sqrt{1+x^2})\\
y'=\frac{1}{1+(x-\sqrt{1+x^2})^2}\times1-{\frac{2x}{2\sqrt{1+x^2}}}\\
y'=\frac{1}{1+(x-\sqrt{1+x^2})^2}\times{\frac{\sqrt{1+x^2}-x}{\sqrt{1+x^2}}}\\
y'=\frac{\sqrt{1+x^2}-x}{(1+(x-\sqrt{1+x^2})^2)(\sqrt{1+x^2})}\\
y'=\frac{1}{2(1+x^2)}$
