University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 159: 85


At $x=2$, $y'=-5$.

Work Step by Step

We have $y=f(g(x))$, $f'(3)=-1$, $g'(2)=5$ and $g(2)=3$. According to the Chain Rule: $$y'=(f(g(x)))'=f'(g(x))\times g'(x)$$ At $x=2$: $$y'(2)=f'(g(2))\times g'(2)$$ We have $g(2)=3$ and $g'(2)=5$: $$y'(2)=f'(3)\times5$$ Finally, we have $f'(3)=-1$: $$y'(2)=-1\times5=-5$$
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