Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 6 - Applications of Integration - 6.8 Logarithmic and Exponential - 6.8 Exercises: 12

Answer

$$\frac{{18x{{\ln }^2}\left( {3{x^2} + 2} \right)}}{{3{x^2} + 2}}$$

Work Step by Step

$$\eqalign{ & \frac{d}{{dx}}\left( {{{\ln }^3}\left( {3{x^2} + 2} \right)} \right) \cr & {\text{use the chain rule}} \cr & = 3{\ln ^2}\left( {3{x^2} + 2} \right)\frac{d}{{dx}}\ln \left( {3{x^2} + 2} \right) \cr & {\text{use the rule }}\frac{d}{{dx}}\left( {\ln u} \right) = \frac{1}{u}\frac{{du}}{{dx}},{\text{ letting }}u = 3{x^2} + 2 \cr & = 3{\ln ^2}\left( {3{x^2} + 2} \right)\left( {\frac{1}{{3{x^2} + 1}}} \right)\frac{d}{{dx}}\left( {3{x^2} + 2} \right) \cr & = 3{\ln ^2}\left( {3{x^2} + 2} \right)\left( {\frac{1}{{3{x^2} + 1}}} \right)\left( {6x} \right) \cr & {\text{simplify}} \cr & = \frac{{18x{{\ln }^2}\left( {3{x^2} + 2} \right)}}{{3{x^2} + 2}} \cr} $$
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