Answer
$y'=(6θ+3secθtanθ)(θ^2+secθ+1)^2$
Work Step by Step
Take the derivative of the equation, and apply Chain Rule to the inner terms:
$y'=3(θ^2+secθ+1)^2\times(2θ+secθtanθ)$
$=(6θ+3secθtanθ)(θ^2+secθ+1)^2$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 3: Derivatives - Practice Exercises - Page 177: 7
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