Thomas' Calculus 13th Edition

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

Chapter 3: Derivatives - Section 3.5 - Derivatives of Trigonometric Functions - Exercises 3.5: 22

Answer

The Derivative is: $\frac{ds}{dt}=\frac{-1}{1-\cos t}$

Work Step by Step

$s=\frac{\sin t}{1-\cos t}$ Applying Derivative rules: $\frac{ds}{dt}=\frac{(1-\cos t)\cdot\frac{d}{dt}(\sin t)-(\sin t)\cdot\frac{d}{dt}(1-\cos t)}{(1-\cos t)^2}$ $\frac{ds}{dt}=\frac{(1-\cos t)(\cos t)-(\sin t)(0+\sin t)}{(1-\cos t)^2}$ $\frac{ds}{dt}=\frac{\cos t-\cos^2t-\sin^2 t}{(1-\cos t)^2}$ $\frac{ds}{dt}=\frac{\cos t-\cos^2t-\sin^2 t}{(1-\cos t)^2}$ $\frac{ds}{dt}=\frac{(-1)(-\cos t+\cos^2t+\sin^2 t)}{(1-\cos t)^2}$ Applying trigonometric identities: $\cos^2 t+\sin^2 t=1$ $\frac{ds}{dt}=\frac{(-1)(1-\cos t)}{(1-\cos t)(1-\cos t)}$ $\frac{ds}{dt}=\frac{-1}{1-\cos t}$
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