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: 37

Answer

$$H'\left( x \right) = 2{\left( {x + 1} \right)^{2x}}\left( {\frac{x}{{x + 1}} + \ln \left( {x + 1} \right)} \right)$$

Work Step by Step

$$\eqalign{ & H\left( x \right) = {\left( {x + 1} \right)^{2x}} \cr & {\text{we use the inverse relationship }}{e^{\ln x}} = x \cr & H\left( x \right) = {e^{\ln {{\left( {x + 1} \right)}^{2x}}}} \cr & H\left( x \right) = {e^{2x\ln \left( {x + 1} \right)}} \cr & {\text{evaluate the derivative}} \cr & H'\left( x \right) = \frac{d}{{dx}}\left( {{e^{2x\ln \left( {x + 1} \right)}}} \right) \cr & {\text{by }}\frac{d}{{dx}}\left( {{e^{u\left( x \right)}}} \right) = {e^{u\left( x \right)}}u'\left( x \right) \cr & H'\left( x \right) = {e^{2x\ln \left( {x + 1} \right)}}\frac{d}{{dx}}\left( {2x\ln \left( {x + 1} \right)} \right) \cr & {\text{product rule}} \cr & H'\left( x \right) = {e^{2x\ln \left( {x + 1} \right)}}\left( {\frac{{2x}}{{x + 1}} + 2\ln \left( {x + 1} \right)} \right) \cr & {\text{simplify}} \cr & H'\left( x \right) = 2{\left( {x + 1} \right)^{2x}}\left( {\frac{x}{{x + 1}} + \ln \left( {x + 1} \right)} \right) \cr} $$
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