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: 8

Answer

The Derivative is: $y'=\csc x - 2\csc^3 x$

Work Step by Step

$y=\frac{\cos x}{sin^{2}x}$ Applying trigonometric identities $y=\frac{\cot x}{\sin x}$ $y=\cot x\csc x$ Applying Derivative Rules: $y'=f'(x)\cdot g(x)+f(x)\cdot g'(x)$ $y'=(-\csc^2 x)(\csc x)+(-\csc x\cot x)(\cot x)$ $y'=-\csc^3 x-\csc x\cot^2 x$ Applying trigonometric identities $\cot^2 x = -1+\csc^2 x$ $y'=-\csc^3 x-\csc x(-1+\csc^2 x)$ $y'=-\csc^3 x + \csc x - \csc^3 x$ $y'=\csc x - 2\csc^3 x$
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