Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - 7.1 Derivative of f(x)=bx and the Number e - Exercises - Page 327: 28


$$ f'(x)= 4(2 e^{3x}+2 e^{-2x})^3(6 e^{3x}-4e^{-2x}).$$

Work Step by Step

Recall that $(e^x)'=e^x$ Recall that $(x^n)'=nx^{n-1}$ Since we have $$ f(x)=(2 e^{3x}+2 e^{-2x})^4$$ then the derivative $ f'(x)$, using the chain rule, is given by $$ f'(x)=4(2 e^{3x}+2 e^{-2x})^3(2 e^{3x}+2 e^{-2x})'=4(2 e^{3x}+2 e^{-2x})^3(6 e^{3x}-4e^{-2x}).$$
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