Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.6 - The Chain Rule - Exercises 3.6 - Page 149: 65

Answer

$-\frac{t\sin (t^2)}{\sqrt{1+cos(t^2)}}$

Work Step by Step

y=$(1+cos(t^2))^{1/2}$ Apply the chain rule: $\frac{dy}{dt}=\frac{1}{2}(1+cos(t^2))^{-1/2}.\frac{d}{dt}1+cos(t^2)$ =$\frac{1}{2}(1+cos(t^2))^{-1/2}(-sin(t^2).\frac{d}{dt}(t^2))$ =$\frac{1}{2}(1+cos(t^2))^{-1/2}(sin(t^2)).2t $ =$-\frac{t\sin (t^2)}{\sqrt{1+cos(t^2)}}$
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