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: 21

Answer

The Derivative is: $\frac{ds}{dt}=\frac{-2\csc t\cot t}{(1-\csc t)^2}$

Work Step by Step

$s=\frac{1\space+\space\csc t}{1\space-\space\csc t}$ Applying Derivative Rules: $y'=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{g^2(x)}$ $\frac{ds}{dt}=\frac{(1-\csc t)\cdot\frac{d}{dt}(1+\csc t)-(1+\csc t)\cdot\frac{d}{dt}(1-\csc t)}{(1-\csc t)^2}$ $\frac{ds}{dt}=\frac{(1-\csc t)(0+(-\csc t\cot t))-(1+\csc t)(0-(-\csc t\cot t))}{(1-\csc t)^2}$ $\frac{ds}{dt}=\frac{-\csc t\cot t+\csc^2 t\cot t-\csc t\cot t-\csc^2t\cot t}{(1-\csc t)^2}$ $\frac{ds}{dt}=\frac{-2\csc t\cot t}{(1-\csc t)^2}$
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