## Thomas' Calculus 13th Edition

$$\frac{{dy}}{{dx}} = \frac{{x\sqrt {{x^2} + 1} }}{{{{\left( {x + 1} \right)}^{2/3}}}}\left( {\frac{1}{x} + \frac{x}{{{x^2} + 1}} - \frac{2}{{3\left( {x + 1} \right)}}} \right)$$
\eqalign{ & y = \frac{{x\sqrt {{x^2} + 1} }}{{{{\left( {x + 1} \right)}^{2/3}}}} \cr & y = \frac{{x{{\left( {{x^2} + 1} \right)}^{1/2}}}}{{{{\left( {x + 1} \right)}^{2/3}}}} \cr & {\text{Take the natural logarithm of both sides:}} \cr & {\text{Use the properties of logarithms}} \cr & \ln y = \ln \frac{{x{{\left( {{x^2} + 1} \right)}^{1/2}}}}{{{{\left( {x + 1} \right)}^{2/3}}}} \cr & {\text{quotient rule:}} \cr & \ln y = \ln x{\left( {{x^2} + 1} \right)^{1/2}} - \ln {\left( {x + 1} \right)^{2/3}} \cr & {\text{product rule:}} \cr & \ln y = \ln x + \ln {\left( {{x^2} + 1} \right)^{1/2}} - \ln {\left( {x + 1} \right)^{2/3}} \cr & {\text{power rule:}} \cr & \ln y = \ln x + \frac{1}{2}\ln \left( {{x^2} + 1} \right) - \frac{2}{3}\ln \left( {x + 1} \right) \cr & {\text{Take derivatives of both sides with respect to }}x \cr & \frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{x} + \frac{1}{2}\left( {\frac{{2x}}{{{x^2} + 1}}} \right) - \frac{2}{3}\left( {\frac{1}{{x + 1}}} \right) \cr & {\text{solve for }}\frac{{dy}}{{dx}} \cr & \frac{{dy}}{{dx}} = y\left( {\frac{1}{x} + \frac{x}{{{x^2} + 1}} - \frac{2}{{3\left( {x + 1} \right)}}} \right) \cr & {\text{substitute }}\frac{{x\sqrt {{x^2} + 1} }}{{{{\left( {x + 1} \right)}^{2/3}}}}{\text{ for }}y{\text{ }} \cr & \frac{{dy}}{{dx}} = \frac{{x\sqrt {{x^2} + 1} }}{{{{\left( {x + 1} \right)}^{2/3}}}}\left( {\frac{1}{x} + \frac{x}{{{x^2} + 1}} - \frac{2}{{3\left( {x + 1} \right)}}} \right) \cr}