Answer
$2e^{2t}\sec^2(1+e^{2t})$
Work Step by Step
$\frac{d}{dt}\tan(1+e^{2t})$
Use chain rule:
$=\sec^2(1+e^{2t})\frac{d}{dt}(1+e^{2t})$
$=\sec^2(1+e^{2t})*e^{2t}*\frac{d}{dt}(2t)$
$=\sec^2(1+e^{2t})*e^{2t}*2$
$=2e^{2t}\sec^2(1+e^{2t})$
Calculus 8th Edition
by Stewart, James
Published by
Cengage
ISBN 10:
1285740629
ISBN 13:
978-1-28574-062-1
Chapter 6 - Inverse Functions - 6.3* The Natural Exponential Function - 6.3* Exercises - Page 453: 44
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