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: 35

Answer

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

Work Step by Step

$$\eqalign{ & y = {\left( {\ln x} \right)^{\ln x}} \cr & {\text{take logarithm natural }} \cr & \ln y = \ln {\left( {\ln x} \right)^{\ln x}} \cr & {\text{logarithm properties}} \cr & \ln y = \ln x\ln \left( {\ln x} \right) \cr & {\text{differentiate}} \cr & \left( {\ln y} \right)' = \left( {\ln x\ln \left( {\ln x} \right)} \right)' \cr & {\text{product rule}} \cr & \frac{{y'}}{y} = \ln x\left( {\ln \left( {\ln x} \right)} \right)' + \ln \left( {\ln x} \right)\left( {\ln x} \right)' \cr & \frac{{y'}}{y} = \ln x\left( {\frac{{\frac{1}{x}}}{{\ln x}}} \right) + \ln \left( {\ln x} \right)\left( {\frac{1}{x}} \right) \cr & \frac{{y'}}{y} = \frac{1}{x} + \frac{{\ln \left( {\ln x} \right)}}{x} \cr & {\text{solve for }}y' \cr & y' = y\left( {\frac{1}{x} + \frac{{\ln \left( {\ln x} \right)}}{x}} \right) \cr & {\text{replace }}y = {\left( {\ln x} \right)^{\ln x}} \cr & y' = {\left( {\ln x} \right)^{\ln x}}\left( {\frac{1}{x} + \frac{{\ln \left( {\ln x} \right)}}{x}} \right) \cr} $$
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