## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 7: Transcendental Functions - Section 7.2 - Natural Logarithms - Exercises 7.2 - Page 381: 34

#### Answer

$$\frac{{dy}}{{dx}} = \frac{5}{{2\left( {x + 1} \right)}} - \frac{{10}}{{x + 2}}$$

#### Work Step by Step

\eqalign{ & y = \ln \sqrt {\frac{{{{\left( {x + 1} \right)}^5}}}{{{{\left( {x + 2} \right)}^{20}}}}} \cr & {\text{rewrite the radical }} \cr & y = \ln {\left( {\frac{{{{\left( {x + 1} \right)}^5}}}{{{{\left( {x + 2} \right)}^{20}}}}} \right)^{1/2}} \cr & {\text{use the power property for logarithms}} \cr & y = \frac{1}{2}\ln \left( {\frac{{{{\left( {x + 1} \right)}^5}}}{{{{\left( {x + 2} \right)}^{20}}}}} \right) \cr & {\text{use the quotient property for logarithms}} \cr & y = \frac{1}{2}\ln {\left( {x + 1} \right)^5} - \frac{1}{2}\ln {\left( {x + 2} \right)^{20}} \cr & y = \frac{5}{2}\ln \left( {x + 1} \right) - \frac{{20}}{2}\ln \left( {x + 2} \right) \cr & y = \frac{5}{2}\ln \left( {x + 1} \right) - 10\ln \left( {x + 2} \right) \cr & {\text{find the derivative of }}y{\text{ with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{5}{2}\frac{d}{{dx}}\left[ {\ln \left( {x + 1} \right)} \right] - 10\frac{d}{{dx}}\left[ {\ln \left( {x + 2} \right)} \right] \cr & {\text{solve the derivatives using }}\frac{d}{{dx}}\left[ {\ln u} \right] = \frac{1}{u}\frac{{du}}{{dx}} \cr & \frac{{dy}}{{dx}} = \frac{5}{2}\left( {\frac{1}{{x + 1}}} \right) - 10\left( {\frac{1}{{x + 2}}} \right) \cr & {\text{simplifying, we get:}} \cr & \frac{{dy}}{{dx}} = \frac{5}{{2\left( {x + 1} \right)}} - \frac{{10}}{{x + 2}} \cr}

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