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: 67

Answer

$$y'=2^x\ln 2$$

Work Step by Step

$$y=2^x$$ Recall Theorem 5: $$\frac{d}{dx}a^u=a^u\ln a\frac{du}{dx}$$ Apply here to find $y'$: $$y'=(2^x)'=2^x\ln 2(x)'$$ $$y'=2^x\ln 2$$
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