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: 40

Answer

$$Q'\left( t \right) = {t^{1/t}}\left( {\frac{{1 - \ln t}}{{{t^2}}}} \right)$$

Work Step by Step

$$\eqalign{ & Q\left( t \right) = {t^{1/t}} \cr & {\text{we use the inverse relationship }}{e^{\ln x}} = x \cr & Q\left( t \right) = {e^{\ln {t^{1/t}}}} \cr & Q\left( t \right) = {e^{\frac{1}{t}\ln t}} \cr & {\text{evaluate the derivative}} \cr & Q'\left( t \right) = \frac{d}{{dt}}\left( {{e^{\frac{1}{t}\ln t}}} \right) \cr & {\text{by }}\frac{d}{{dx}}\left( {{e^{u\left( x \right)}}} \right) = {e^{u\left( x \right)}}u'\left( x \right) \cr & Q'\left( t \right) = {e^{\frac{1}{t}\ln t}}\frac{d}{{dt}}\left( {\frac{1}{t}\ln t} \right) \cr & {\text{product rule}} \cr & Q'\left( t \right) = {e^{\frac{1}{t}\ln t}}\left( {\frac{1}{t}\left( {\frac{1}{t}} \right) - \frac{{\ln t}}{{{t^2}}}} \right) \cr & {\text{simplify}} \cr & Q'\left( t \right) = {t^{1/t}}\left( {\frac{{1 - \ln t}}{{{t^2}}}} \right) \cr} $$
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