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: 50

Answer

$$2{x^{2x}}\left( {\ln x + 1} \right)$$

Work Step by Step

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