Chain rule with fraction In the case of $$f(x)=\ln\big(x+\sqrt{1+x^2}\big)$$
in the derivative we multiply $$f'(x)=\frac{1}{x+\sqrt{1+x^2}}\bigg(1+\frac{2x}{2\sqrt{1+x^2}}\bigg)$$ when the expression multiply the numerator?   

Following up on the tip of exponentiation of both side.
$$
\mathrm{e}^{f(x)} = \mathrm{e}^{\ln\left(x+\sqrt{1+x^2}\right)} = x+\sqrt{1+x^2}
$$
taking the derivative of both sides
$$
\mathrm{e}^{f(x)} f'(x) = 1 + \frac{1}{2}\frac{2x}{\sqrt{1+x^2}} = 1+\frac{x}{\sqrt{1+x^2}} = \frac{\sqrt{1+x^2} + x}{\sqrt{1+x^2}}
$$
bow the l.h.s we have
$$
\mathrm{e}^{f(x)} f'(x)  = \left(x+\sqrt{1+x^2}\right)\cdot f'(x)
$$
equating the last term in the previous two equations we get
$$
\left(x+\sqrt{1+x^2}\right)\cdot f'(x) = \left(x+\sqrt{1+x^2}\right)\cdot \frac{1}{\sqrt{1+x^2}}
$$
thus we get
$$
f'(x) = \frac{1}{\sqrt{1+x^2}}
$$
the trick used here reduces the pain of doing derivations of the form
$$
f(x) = g(u(x))
$$
where the inverse of $g$ yields forms that are easier to handle in the differentiation such as $\ln (x) $ and $\mathrm{e}^x$.
