Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.9 Derivatives of Logarithmic and Exponential Functions - 3.9 Exercises - Page 212: 75

Answer

$$\frac{{dy}}{{dx}} = {3^x}\left( {\ln 3} \right)$$

Work Step by Step

$$\eqalign{ & y = {3^x} \cr & \left( i \right){\text{Using the fact that }}{b^x} = {e^{x\ln b}},{\text{ then}} \cr & y = {e^{x\ln 3}} \cr & {\text{Differentiating}} \cr & \frac{{dy}}{{dx}} = {e^{x\ln 3}}\frac{d}{{dx}}\left[ {x\ln 3} \right] \cr & \frac{{dy}}{{dx}} = {e^{x\ln 3}}\left( {\ln 3} \right) \cr & {\text{Substitute }}{3^x}{\text{ for }}{e^{x\ln 3}} \cr & \frac{{dy}}{{dx}} = {3^x}\left( {\ln 3} \right) \cr & \cr & \left( {ii} \right){\text{Using the logarithmic differentiation}} \cr & \ln y = \ln {3^x} \cr & \ln y = x\ln 3 \cr & {\text{Differentiate both sides}} \cr & \frac{1}{y}\frac{{dy}}{{dx}} = \ln 3 \cr & {\text{Solving for }}\frac{{dy}}{{dx}} \cr & \frac{{dy}}{{dx}} = y\ln 3 \cr & {\text{Where }}y = {3^x} \cr & \frac{{dy}}{{dx}} = {3^x}\left( {\ln 3} \right) \cr} $$
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