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: 11

Answer

The Derivative is: $y'=\frac{-\csc^2 x}{(1+\cot x)^2}$ OR $y'=-\frac{1}{1+\sin(2x)}$

Work Step by Step

$y=\frac{\cot x}{1+\cot x}$ $y'=\frac{(-\csc^2 x)(1+\cot x)-(\cot x)(-\csc^2 x)}{(1+\cot x)^2}$ $y'=\frac{-\csc^2 x-\frac{\cos x}{\sin^3 x}+\frac{\cos x}{\sin^3 x}}{(1+\cot x)^2}$ $y'=\frac{-\csc^2 x}{(1+\cot x)^2}$ $y'=\frac{-\frac{1}{\sin^2 x}}{1+2\frac{\cos x}{\sin x}+\frac{\cos^2 x}{\sin^2 x}}$ $y'=\frac{-\frac{1}{\sin^2 x}}{\frac{\sin^2 x+2\sin x\cos x+\cos^2 x}{\sin^2 x}}$ $y'=-\frac{1}{\sin^2 x}\frac{\sin^2 x}{\sin^2 x+\cos^2 x+2\sin x\cos x}$ $y'=-\frac{1}{\sin^2 x+\cos^2 x+2\sin x\cos x}$ Applying trigonometric Identities $2\sin x\cos x =\sin(2x)$ and $\sin^2 x+\cos^2 x=1$ $y'=-\frac{1}{1+\sin(2x)}$
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