Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.3 Logarithms and Their Derivatives - Exercises - Page 343: 77


$$ y'=x^{3^x}( 3^x\ln3\ln x+\frac{3^x}{x}).$$

Work Step by Step

Recall that $(e^x)'=e^x $, $(\ln x)'=\dfrac{1}{x}$. We have $$ f(x)=x^{3^x}=e^{ 3^x\ln x}.$$ Now taking the derivative, we get $$ f'(x)= e^{ 3^x\ln x}( 3^x\ln x)'=e^{ 3x\ln x}( 3^x\ln3\ln x+\frac{3^x}{x})\\ =x^{3^x}( 3^x\ln3\ln x+\frac{3^x}{x}).$$
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