University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.8 - Derivatives of Inverse Functions and Logarithms - Exercises - Page 175: 24


$$y'=4x(x^2\ln x)^3(2\ln x+1)$$

Work Step by Step

$$y=(x^2\ln x)^4$$ Recall the following Derivative Rules: $$\frac{d}{dx}(\ln x)=\frac{1}{x}$$ Therefore, we have $$y'=\Big((x^2\ln x)^4\Big)'=4(x^2\ln x)^3(x^2\ln x)'$$ $$y'=4(x^2\ln x)^3\Big((x^2)'\ln x+x^2(\ln x)'\Big)$$ $$y'=4(x^2\ln x)^3\Big(2x\ln x+x^2\times\frac{1}{x}\Big)$$ $$y'=4(x^2\ln x)^3(2x\ln x+x)$$ $$y'=4x(x^2\ln x)^3(2\ln x+1)$$
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