Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.2 Exponential Functions and Their Derivatives - 6.2 Exercises - Page 419: 45


$F'(t)=e^{tsin2t}.[ 2t cos(2t)+sin 2t]$

Work Step by Step

Need to solve $F'(t)=\frac{d}{dt}(e^{tsin2t)}$ Apply chain rule. Suppose $u = tsin2t$ Then, $F'(t)=\frac{d}{dt}(e^{u})$ Also we can write as follows: $F'(t)=\frac{dF}{du}\times\frac{du}{dF}$ First determine $\frac{dF}{du}$ $F'(t)=\frac{dF}{dt}=\frac{dF}{du}\times\frac{du}{dt}$ ...(1) Thus, $\frac{dF}{du}=e^{u}$ and $\frac{du}{dt} =2t cos2t+sin2t$ From equation (1) , we have $F'(t)=\frac{dF}{dt}=\frac{dF} {du}\times\frac{du}{dt}$ $F'(t)=e^{u}.[ 2t cos(2t)+sin 2t]$ Hence, $F'(t)=e^{tsin2t}.[ 2t cos(2t)+sin 2t]$
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