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: 69


$$y'=\frac{5^{\sqrt s}\ln5}{2\sqrt s}$$

Work Step by Step

$$y=5^{\sqrt s}$$ Recall Theorem 5: $$\frac{d}{ds}a^u=a^u\ln a\frac{du}{ds}$$ Apply here to find $y'$: $$y'=(5^{\sqrt s})'=5^{\sqrt s}\ln 5(\sqrt s)'=5^{\sqrt s}\ln 5\times\frac{1}{2\sqrt s}$$ $$y'=\frac{5^{\sqrt s}\ln5}{2\sqrt s}$$
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