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: 63

Answer

(a) $$h'(1)=30$$ (b) $$H'(1)=36$$

Work Step by Step

(a) $$h(x)=f(g(x))$$ The derivative of $h$ would be $$h'(x)=\frac{df(g(x))}{dx}$$ Apply the Chain Rule, we have $$h'(x)=\frac{df(g(x))}{dg(x)}\frac{dg(x)}{dx}$$ $$h'(x)=f'(g(x))g'(x)$$ Therefore, $$h'(1)=f'(g(1))g'(1)$$ From the table, we see that $g(1)=2$ and $g'(1)=6$ $$h'(1)=f'(2)\times6$$ Again, from the table, $f'(2)=5$ $$h'(1)=5\times6=30$$ (b) $$H(x)=g(f(x))$$ $$H'(x)=\frac{dg(f(x))}{dx}$$ Apply the Chain Rule: $$H'(x)=\frac{dg(f(x))}{df(x)}\frac{df(x)}{dx}$$ $$H'(x)=g'(f(x))f'(x)$$ So, $$H'(1)=g'(f(1))f'(1)$$ From the table, $f(1)=3$ and $f'(1)=4$ $$H'(1)=g'(3)\times4$$ From the table, $g'(3)=9$ $$H'(1)=9\times4=36$$
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