Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.3 - Exponential Functions - Exercises 7.3 - Page 390: 6

Answer

$$\frac{{dy}}{{dx}} = \frac{2}{3}{e^{2x/3}}$$

Work Step by Step

$$\eqalign{ & y = {e^{2x/3}} \cr & {\text{Find the derivative of }}y{\text{ with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{e^{2x/3}}} \right] \cr & {\text{We can use the formula }}\frac{d}{{dx}}{e^u} = {e^u}\frac{{du}}{{dx}}{\text{ where }}u{\text{ is any differentiable function of }}x. \cr & {\text{For this exercise you can note that }}u = \frac{{2x}}{3},{\text{ then}} \cr & \frac{{dy}}{{dx}} = {e^{2x/3}}\frac{d}{{dx}}\left[ {\frac{{2x}}{3}} \right] \cr & {\text{solve the derivative and simplify}} \cr & \frac{{dy}}{{dx}} = {e^{2x/3}}\left( {\frac{2}{3}} \right) \cr & \frac{{dy}}{{dx}} = \frac{2}{3}{e^{2x/3}} \cr} $$
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