Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises - Page 205: 74

Answer

$F'(1) = 198$

Work Step by Step

$F(x) = f(xf(xf(x)))$ $F'(x) = f'(xf(xf(x)))\cdot\frac{d}{dx}(xf(xf(x)))$ $F'(x) = f'(xf(xf(x)))\cdot~[f(xf(x))+x~f'(xf(x))*(xf(x))']$ $F'(x) = f'(xf(xf(x)))\cdot~[f(xf(x))+x~f'(xf(x))*(xf'(x)+1f(x))]$ $F'(1) = f'(1f(1f(1)))\cdot~[f(1f(1))+1~f'(1f(1))*(1*f'(1)+1*f(1))]$ $F'(1) = f'(f(f(1)))\cdot~[f(f(1))+f'(f(1))*(4+2)]$ $F'(1) = f'(f(2))\cdot~[f(2)+f'(2)*(4+2)]$ $F'(1) = f'(3)\cdot~[3+5*6]$ $F'(1) = (6)\cdot~(33)$ $F'(1) = 198$
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