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 174: 14

Answer

$$y'=\frac{3}{2t}+\frac{1}{2\sqrt t}$$

Work Step by Step

$$y=\ln(t^{3/2})+\sqrt t=\ln(t^{3/2})+t^{1/2}$$ Recall the Derivative Rule for Natural Logarithm: $$\frac{d}{dx}(\ln u)=\frac{1}{u}\frac{du}{dx}$$ Therefore, we have $$y'=(\ln(t^{3/2})+t^{1/2})'=\frac{1}{t^{3/2}}(t^{3/2})'+\frac{1}{2}t^{-1/2}$$ $$y'=\frac{1}{t^{3/2}}\frac{3}{2}t^{1/2}+\frac{1}{2t^{1/2}}=\frac{3t^{1/2}}{2t^{3/2}}+\frac{1}{2\sqrt t}$$ $$y'=\frac{3}{2t}+\frac{1}{2\sqrt t}$$
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