## Calculus (3rd Edition)

$$y'= \frac{x }{\sqrt{1+x^2}\sqrt{2+x^2}}.$$
Since $y=\sinh^{-1} (\sqrt{x^2+1})$, then the derivative, by using the chain rule and the fact that $\frac{d}{d x} \sinh ^{-1} u=\frac{u'}{\sqrt{u^{2}+1}}$, is given by $$y'= \frac{1}{\sqrt{1+(\sqrt{x^2+1})^2}}(\sqrt{1+x^2})'=\frac{2x }{2\sqrt{1+x^2}\sqrt{2+x^2}}\\ =\frac{x }{\sqrt{1+x^2}\sqrt{2+x^2}}.$$