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: 87



Work Step by Step

$$y=\log_2(8t^{\ln2})$$ Recall Theorem 7: $$\frac{d}{dt}\log_au=\frac{1}{u\ln a}\frac{du}{dt}$$ So the derivative of $y$ is: $$y'=\frac{1}{8t^{\ln2}\ln2}(8t^{\ln2})'$$ Remember that $(x^n)'=nx^{n-1}$ $$y'=\frac{(8\ln2)t^{(\ln2)-1}}{8t^{\ln2}\ln2}=\frac{t^{(\ln2)-1}}{t^{\ln2}}$$ $$y'=t^{\ln2-1-\ln2}=t^{-1}$$ $$y'=\frac{1}{t}$$
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