Answer
$$\frac{{dy}}{{dx}} = \frac{{{x^3}}}{{\sqrt {{x^2} + 1} }}\left( {\frac{3}{x} - \frac{x}{{{x^2} + 1}}} \right)$$
Work Step by Step
$$\eqalign{
& y = \frac{{{x^3}}}{{\sqrt {{x^2} + 1} }} \cr
& {\text{Take natural logarithm on both sides}} \cr
& \ln y = \ln \frac{{{x^3}}}{{\sqrt {{x^2} + 1} }} \cr
& {\text{Rewrite the radical}} \cr
& \ln y = \ln \frac{{{x^3}}}{{{{\left( {{x^2} + 1} \right)}^{1/2}}}} \cr
& {\text{Using the logarithmic properties}} \cr
& \ln y = \ln {x^3} - \ln {\left( {{x^2} + 1} \right)^{1/2}} \cr
& \ln y = 3\ln x - \frac{1}{2}\ln \left( {{x^2} + 1} \right) \cr
& {\text{Differentiate both sides with respect to }}x \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = 3\left( {\frac{1}{x}} \right) - \frac{1}{2}\left( {\frac{{2x}}{{{x^2} + 1}}} \right) \cr
& {\text{Multiply}} \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \frac{3}{x} - \frac{x}{{{x^2} + 1}} \cr
& {\text{Solving for }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = y\left( {\frac{3}{x} - \frac{x}{{{x^2} + 1}}} \right) \cr
& {\text{Substituting }}y = \frac{{{x^3}}}{{\sqrt {{x^2} + 1} }} \cr
& \frac{{dy}}{{dx}} = \frac{{{x^3}}}{{\sqrt {{x^2} + 1} }}\left( {\frac{3}{x} - \frac{x}{{{x^2} + 1}}} \right) \cr} $$
