Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.3 - Derivatives of Trigonometric Functions - 3.3 Exercises - Page 196: 17


$\dfrac{d}{dx}(\csc x)=-\csc x\cot x$

Work Step by Step

$\dfrac{d}{dx}(\csc x)$ We know that $\csc x=\dfrac{1}{\sin x}$, so we make that substitution and differentiate applying the quotient rule: $\dfrac{d}{dx}(\dfrac{1}{\sin x})=\dfrac{(\sin x)(1)'-(1)(\sin x)'}{(\sin x)^{2}}=...$ $...=\dfrac{(\sin x)(0)-(1)(\cos x)}{\sin^{2}x}=\dfrac{-\cos x}{\sin^{2}x}$ This is the derivative. of $\csc x$. Since we have two factors of sine in the denominator, let's write this expression as a product of two fractions: $\dfrac{-\cos x}{\sin^{2}x}=(\dfrac{1}{\sin x})(-\dfrac{\cos x}{\sin x})$ Since $\dfrac{1}{\sin x}=\csc x$ and $\dfrac{\cos x}{\sin x}=\cot x$, we finally get: $\dfrac{d}{dx}(\csc x)=-\csc x\cot x$
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