Answer
$$\frac{x}{{\sqrt {{x^2} + 1} }}$$
Work Step by Step
$$\eqalign{
& differentiate \cr
& = \frac{d}{{dx}}\left( {\sqrt {{x^2} + 1} + C} \right) \cr
& = \frac{d}{{dx}}\left( {\sqrt {{x^2} + 1} } \right) + \frac{d}{{dx}}\left( C \right) \cr
& {\text{rewritting the radical}} \cr
& = \frac{d}{{dx}}\left( {{{\left( {{x^2} + 1} \right)}^{1/2}}} \right) + \frac{d}{{dx}}\left( C \right) \cr
& {\text{by the chain rule}} \cr
& = \frac{1}{2}{\left( {{x^2} + 1} \right)^{ - 1/2}}\frac{d}{{dx}}\left( {{x^2} + 1} \right) + \frac{d}{{dx}}\left( C \right) \cr
& = \frac{1}{2}{\left( {{x^2} + 1} \right)^{ - 1/2}}\left( {2x} \right) + 0 \cr
& {\text{simplify}} \cr
& = \frac{x}{{\sqrt {{x^2} + 1} }} \cr} $$
