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 158: 66


$$\frac{dy}{dt}=\frac{\cos\sqrt{(1+\sqrt t)}}{\sqrt t\sqrt{1+\sqrt t}}$$

Work Step by Step

$$\frac{dy}{dt}=\frac{d}{dt}4\sin\sqrt{1+\sqrt t}=\frac{d}{dt}4\sin(1+t^{1/2})^{1/2}$$ Following the Chain Rule: $$\frac{dy}{dt}=4\cos(1+t^{1/2})^{1/2}\frac{d}{dt}(1+t^{1/2})^{1/2}$$ $$\frac{dy}{dt}=4\cos(1+t^{1/2})^{1/2}\frac{1}{2}(1+t^{1/2})^{-1/2}\frac{d}{dt}(1+t^{1/2})$$ $$\frac{dy}{dt}=2\cos(1+t^{1/2})^{1/2}(1+t^{1/2})^{-1/2}\Big(\frac{1}{2}t^{-1/2}\Big)$$ $$\frac{dy}{dt}=\cos(1+t^{1/2})^{1/2}(1+t^{1/2})^{-1/2}t^{-1/2}$$ $$\frac{dy}{dt}=\frac{\cos\sqrt{(1+\sqrt t)}}{\sqrt t\sqrt{1+\sqrt t}}$$
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