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: 26


$$\frac{ds}{dt}=\frac{\sin t+\cos t +1}{2(1+\cos t)\sqrt{(1+\sin t)(1+\cos t)}}$$

Work Step by Step

$$s(t)=\sqrt{\frac{1+\sin t}{1+\cos t}}$$ Then $$\frac{ds}{dt}=\frac{d\sqrt{\frac{1+\sin t}{1+\cos t}}}{dt}$$ Let $u=\frac{1+\sin t}{1+\cos t}$, then $s(u)=\sqrt u$. According to the Chain Rule: $$\frac{ds}{dt}=\frac{ds}{du}\frac{du}{dt}$$ $$\frac{ds}{dt}=\frac{d(\sqrt u)}{du}\frac{d(\frac{1+\sin t}{1+\cos t})}{dt}$$ *Find $\frac{d(\sqrt u)}{du}$ $$\frac{d(\sqrt u)}{du}=\frac{d(u^{1/2})}{du}=\frac{1}{2}u^{-1/2}$$ *Find $\frac{d(\frac{1+\sin t}{1+\cos t})}{dt}$ $$\frac{d(\frac{1+\sin t}{1+\cos t})}{dt}=\Bigg(\frac{1+\sin t}{1+\cos t}\Bigg)'$$ $$\frac{d(\frac{1+\sin t}{1+\cos t})}{dt}=\frac{(1+\sin t)'(1+\cos t)-(1+\sin t)(1+\cos t)'}{(1+\cos t)^2}$$ $$\frac{d(\frac{1+\sin t}{1+\cos t})}{dt}=\frac{\cos t(1+\cos t)-(1+\sin t)(-\sin t)}{(1+\cos t)^2}$$ $$\frac{d(\frac{1+\sin t}{1+\cos t})}{dt}=\frac{\cos t+\cos^2 t+\sin t+\sin^2 t}{(1+\cos t)^2}$$ $$\frac{d(\frac{1+\sin t}{1+\cos t})}{dt}=\frac{\cos t+\sin t+1}{(1+\cos t)^2}$$ Therefore, $$\frac{ds}{dt}=\frac{1}{2}u^{-1/2}\times\frac{\cos t+\sin t+1}{(1+\cos t)^2}$$ $$\frac{ds}{dt}=\frac{1}{2}(\frac{1+\sin t}{1+\cos t})^{-1/2}\times\frac{\cos t+\sin t+1}{(1+\cos t)^2}$$ $$\frac{ds}{dt}=\frac{(1+\sin t)^{-1/2}(\sin t+\cos t+1)}{2(1+\cos t)^{3/2}}$$ $$\frac{ds}{dt}=\frac{\sin t+\cos t +1}{2\sqrt{1+\sin t}\sqrt{(1+\cos t)^3}}$$ $$\frac{ds}{dt}=\frac{\sin t+\cos t +1}{2(1+\cos t)\sqrt{(1+\sin t)(1+\cos t)}}$$
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