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: 47



Work Step by Step

$ q=sin(\frac{t}{\sqrt{t+1}})$ Apply the chain rule: $\frac{dq}{dt} =cos(\frac{t}{\sqrt{t+1}}).\frac{d}{dt}(\frac{t}{\sqrt{t+1}})$ =cos $(\frac{t}{\sqrt{t+1}}).\frac{\sqrt{t+1} (1)-t\frac{d}{dt}\sqrt{t+1}}{(\sqrt{t+1})^2}$ =cos $(\frac{1}{\sqrt{t+1}}).\frac{\sqrt{t+1}.\frac{1}{2\sqrt{t+1}}}{t+1}$ =cos $(\frac{t}{\sqrt{t+1}})(\frac{2(t+1)-t}{2(t+1)^{3/2}})$ =$(\frac{t+2}{2(t+1)^{3/2}})cos(\frac{t}{\sqrt{t+1}})$
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