A surprisingly simple derivative

Domain. Since $\sqrt{1+x^2}>|x|\ge-x$, we have $u(x)=x+\sqrt{1+x^2}>0$ for every $x$, so $f$ is well defined and differentiable on $\mathbb{R}$.

Derivative of $u$. $u'(x)=1+\dfrac x{\sqrt{1+x^2}}=\dfrac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}=\dfrac{u(x)}{\sqrt{1+x^2}}$.

Chain rule.
$$f'(x)=\frac{u'(x)}{u(x)}=\frac1{\sqrt{1+x^2}}.$$

Remark. This function is the inverse hyperbolic sine $\operatorname{arsinh}$, and the result explains why $\int\frac{dx}{\sqrt{1+x^2}}=\ln\left(x+\sqrt{1+x^2}\right)$.