Answer
$f'(g(x)) = 2cos(2x+1)$
Work Step by Step
$f(u) = sin(u), g(x) = 2x+1$
$f(g(x)) = sin(2x+1)$
$f'(u) = cos(u)$ and $g'(x) = 2$
$f'(g(x)) = g'(x)f'(u)$
$f'(g(x)) = 2(cos(u))$
We know that $u = g(x) = 2x+1$ so:
$f'(g(x)) = 2cos(2x+1)$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 3 - Differentiation - 3.7 The Chain Rule - Exercises - Page 146: 23
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