#### Answer

$y' = sinx +tanx\ secx=sinx(sec^2{x}+1)$

#### Work Step by Step

$y = sinx\ tanx$
$y' = cosx\ tanx + sinx\ sec^{2}x$
$y' = cosx\ \frac{sinx}{cosx} + sinx\ \frac{1}{cos^{2}x}$
$y' = sinx +tanx\ secx$

Published by
Pearson

ISBN 10:
0321999584

ISBN 13:
978-0-32199-958-0

$y' = sinx +tanx\ secx=sinx(sec^2{x}+1)$

You can help us out by revising, improving and updating this answer.

Update this answerAfter you claim an answer you’ll have **24 hours** to send in a draft. An editor
will review the submission and either publish your submission or provide feedback.