Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.5 The Chain Rule - 2.5 Exercises - Page 158: 30


$$s'(t)=\frac{\sin t+\cos t+1}{2\sqrt{1+\sin t}(1+\cos t)^{3/2}}$$

Work Step by Step

$$s'(t)=\left(\sqrt{\frac{1+\sin t}{1+\cos t}}\right)'=\frac{1}{2\sqrt{\frac{1+\sin t}{1+\cos t}}}\left(\frac{1+\sin t}{1+\cos t}\right)'= \frac{\sqrt{1+\cos t}}{2\sqrt{1+\sin t}}\frac{(1+\sin t)'(1+\cos t)-(1+\sin t)(1+\cos t)'}{(1+\cos t)^2}= \frac{\sqrt{1+\cos t}}{2\sqrt{1+\sin t}}\frac{\cos t(1+\cos t)-(1+\sin t)(-\sin t)}{(1+\cos t)^2}= \frac{\sqrt{1+\cos t}}{2\sqrt{1+\sin t}}\frac{\cos t+\cos^2t+\sin t+\sin^2t}{(1+\cos t)^2}= \frac{\sqrt{1+\cos t}(\sin t+\cos t+1)}{2\sqrt{1+\sin t}(1+\cos t)^2}= \frac{\sin t+\cos t+1}{2\sqrt{1+\sin t}(1+\cos t)^{3/2}}$$
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