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: 73

Answer

$$F'(0)=96$$

Work Step by Step

$$F(x)=f(3f(4f(x))))$$ $$F'(x)=\frac{df(3f(4f(x))))}{dx}$$ According to Chain Rule: $$F'(x)=\frac{df(3f(4f(x))))}{d(3f(4f(x)))}\frac{3d(f(4f(x)))}{d(4f(x))}\frac{4df(x)}{dx}$$ $$F'(x)=f'(3f(4f(x)))\times3f'(4f(x))\times4f'(x)$$ $$F'(x)=12f'(3f(4f(x)))f'(4f(x))f'(x)$$ Therefore, $$F'(0)=12f'(3f(4f(0)))f'(4f(0))f'(0)$$ We know $f(0)=0$ and $f'(0)=2$ $$F'(0)=12f'(3f(4\times0))f'(4\times0)\times2$$ $$F'(0)=24f'(3f(0))f'(0)$$ Again, apply $f(0)=0$ and $f'(0)=2$ $$F'(0)=24f'(3\times0)\times2$$ $$F'(0)=48f'(0)$$ Finally, apply $f'(0)=2$ $$F'(0)=48\times2=96$$
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