Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.3 Derivatives Of Inverse Functions; Derivatives And Integrals Involving Exponential Functions - Exercises Set 6.3 - Page 432: 32

Answer

$$y' = {x^{\sin x - 1}}\sin x + {x^{\sin x}}\ln x\left( {\cos x} \right)$$

Work Step by Step

$$\eqalign{ & y = {x^{\sin x}} \cr & {\text{take logarithm natural on both sides}} \cr & \ln y = \ln {x^{\sin x}} \cr & {\text{logarithm properties}} \cr & \ln y = \sin x\ln x \cr & {\text{differentiate}} \cr & \left( {\ln y} \right)' = \left( {\sin x\ln x} \right)' \cr & {\text{product rule}} \cr & \frac{{y'}}{y} = \sin x\left( {\ln x} \right)' + \ln x\left( {\sin x} \right)' \cr & \frac{{y'}}{y} = \sin x\left( {\frac{1}{x}} \right) + \ln x\left( {\cos x} \right) \cr & y' = y\left[ {\frac{{\sin x}}{x} + \ln x\left( {\cos x} \right)} \right] \cr & {\text{replace }}y = {x^{\sin x}} \cr & y' = {x^{\sin x}}\left[ {\frac{{\sin x}}{x} + \ln x\left( {\cos x} \right)} \right] \cr & {\text{simplify}} \cr & y' = {x^{\sin x - 1}}\sin x + {x^{\sin x}}\ln x\left( {\cos x} \right) \cr} $$
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