Answer
See proof
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\ln \left( {x + \sqrt {{x^2} + 1} } \right) = \frac{1}{{\sqrt {{x^2} + 1} }} \cr
& {\text{Differentiating the left side}}{\text{, use }}\frac{d}{{dx}}\left[ {\ln u} \right] = \frac{1}{u}\frac{{du}}{{dx}} \cr
& \frac{d}{{dx}}\ln \left( {x + \sqrt {{x^2}} + 1} \right) = \frac{1}{{x + \sqrt {{x^2} + 1} }}\frac{d}{{dx}}\left[ {x + \sqrt {{x^2} + 1} } \right] \cr
& {\text{Computing the derivatives}} \cr
& = \frac{1}{{x + \sqrt {{x^2} + 1} }}\left( {1 + \frac{{2x}}{{2\sqrt {{x^2} + 1} }}} \right) \cr
& = \frac{1}{{x + \sqrt {{x^2} + 1} }}\left( {1 + \frac{x}{{\sqrt {{x^2} + 1} }}} \right) \cr
& {\text{Simplifying}} \cr
& = \frac{1}{{x + \sqrt {{x^2} + 1} }}\left( {\frac{{\sqrt {{x^2} + 1} + x}}{{\sqrt {{x^2} + 1} }}} \right) \cr
& = \frac{1}{{\sqrt {{x^2} + 1} }} \cr
& {\text{Therefore}}{\text{, the statement has been verified.}} \cr} $$
