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: 39

Answer

$$G'\left( y \right) = {y^{\sin y}}\left( {\frac{{\sin y}}{y} + \ln y\cos y} \right)$$

Work Step by Step

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