Calculus (3rd Edition)

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



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 $x=1$, $(f(2x+g(x)))'=$ $f'(2\times1+g(1))\times(2+g'(1))$ $=f'(2+4)\times(2+5)=4\times7=28$
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