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: 64

Answer

(a) $$F'(2)=20$$ (b) $$G'(3)=63$$

Work Step by Step

(a) $$F(x)=f(f(x))$$ The derivative: $$F'(x)=\frac{df(f(x))}{dx}$$ Apply the Chain Rule, we have $$F'(x)=\frac{df(f(x))}{df(x)}\frac{df(x)}{dx}$$ $$F'(x)=f'(f(x))f'(x)$$ Therefore, $$F'(2)=f'(f(2))f'(2)$$ From the table, we see that $f(2)=1$ and $f'(2)=5$. So, $$F'(2)=f'(1)\times5$$ Again, from the table, $f'(1)=4$. Therefore, $$F'(2)=4\times5=20$$ (b) $$G(x)=g(g(x))$$ The derivative: $$G'(x)=\frac{dg(g(x))}{dx}$$ Apply the Chain Rule: $$G'(x)=\frac{dg(g(x))}{dg(x)}\frac{dg(x)}{dx}$$ $$G'(x)=g'(g(x))g'(x)$$ So, $$G'(3)=g'(g(3))g'(3)$$ From the table, $g(3)=2$ and $g'(3)=9$ $$G'(3)=g'(2)\times9$$ From the table, $g'(2)=7$ $$G'(3)=7\times9=63$$
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