## Thomas' Calculus 13th Edition

$$\frac{{dy}}{{dx}} = \frac{{5{{\left( {x + 1} \right)}^5}}}{{{{\left( {2x + 1} \right)}^{5/2}}}}\left( {\frac{1}{{x + 1}} - \frac{1}{{2x + 1}}} \right)$$
\eqalign{ & y = \sqrt {\frac{{{{\left( {x + 1} \right)}^{10}}}}{{{{\left( {2x + 1} \right)}^5}}}} \cr & y = \frac{{{{\left( {x + 1} \right)}^5}}}{{{{\left( {2x + 1} \right)}^{5/2}}}} \cr & {\text{Take the natural logarithm of both sides:}} \cr & {\text{Use the properties of logarithms}} \cr & \ln y = \ln \frac{{{{\left( {x + 1} \right)}^5}}}{{{{\left( {2x + 1} \right)}^{5/2}}}} \cr & {\text{quotient rule:}} \cr & \ln y = \ln {\left( {x + 1} \right)^5} - \ln {\left( {2x + 1} \right)^{5/2}} \cr & {\text{power rule:}} \cr & \ln y = 5\ln \left( {x + 1} \right) - \frac{5}{2}\ln \left( {2x + 1} \right) \cr & {\text{Take derivatives of both sides with respect to }}x \cr & \frac{1}{y}\frac{{dy}}{{dx}} = 5\left( {\frac{1}{{x + 2}}} \right) - \frac{5}{2}\left( {\frac{2}{{2x + 1}}} \right) \cr & {\text{solve for }}\frac{{dy}}{{dx}} \cr & \frac{{dy}}{{dx}} = y\left( {\frac{5}{{x + 1}} - \frac{5}{{2x + 1}}} \right) \cr & {\text{substitute }}\frac{{{{\left( {x + 1} \right)}^5}}}{{{{\left( {2x + 1} \right)}^{5/2}}}}{\text{ for }}y{\text{ }} \cr & \frac{{dy}}{{dx}} = \frac{{5{{\left( {x + 1} \right)}^5}}}{{{{\left( {2x + 1} \right)}^{5/2}}}}\left( {\frac{1}{{x + 1}} - \frac{1}{{2x + 1}}} \right) \cr}