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: 50

Answer

$f'(t)=sin2t(e^{sin^{2}t})[sin(2e^{sin^{2}t})]$

Work Step by Step

$f'(t)=\frac{d}{dt}sin^{2}(e^{sin^{2}t})$ $=2sin(e^{sin^{2}t})\frac{d}{dt}[sin(e^{sin^{2}t})]$ $=2sin(e^{sin^{2}t})cos(e^{sin^{2}t}).(e^{sin^{2}t})\frac{d}{dt}({sin^{2}t)}$ $=sin2(e^{sin^{2}t})(e^{sin^{2}t})(2sintcost)$ Hence,$f'(t)=sin2t(e^{sin^{2}t})[sin(2e^{sin^{2}t})]$
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