Answer
28
Work Step by Step
$\frac{d}{dx}(f(2x+g(x)))=f'(2x+g(x))\times(2x+g(x))'$
$=f'(2x+g(x))\times(2+g'(x))$
When $x=1$,
$(f(2x+g(x)))'=$
$f'(2\times1+g(1))\times(2+g'(1))$
$=f'(2+4)\times(2+5)=4\times7=28$
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 147: 90
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