University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.5 - Derivatives of Trigonometric Functions - Exercises - Page 152: 60


$(\sec x)'=\sec x\tan x$ $(\csc x)'=-\csc x\cot x$ $(\cot x)'=-\csc^2x$

Work Step by Step

Find the derivative of a) $\sec x$: $$(\sec x)'=\Big(\frac{1}{\cos x}\Big)'=\frac{(1)'\cos x-1(\cos x)'}{\cos^2x}=\frac{0\times\cos x-(-\sin x)}{\cos^2x}$$ $$(\sec x)'=\frac{\sin x}{\cos^2x}=\frac{1}{\cos x}\times\frac{\sin x}{\cos x}=\sec x\tan x$$ b) $\csc x$: $$(\csc x)'=\Big(\frac{1}{\sin x}\Big)'=\frac{(1)'\sin x-1(\sin x)'}{\sin^2x}=\frac{0\times\sin x-\cos x}{\sin^2x}$$ $$(\csc x)'=\frac{-\cos x}{\sin^2x}=-\frac{1}{\sin x}\times\frac{\cos x}{\sin x}=-\csc x\cot x$$ c) $\cot x$: $$(\cot x)'=\Big(\frac{\cos x}{\sin x}\Big)'=\frac{(\cos x)'\sin x-\cos x(\sin x)'}{\sin^2x}$$ $$(\cot x)'=\frac{-\sin x\sin x-\cos x\cos x}{\sin^2x}=-\frac{\sin^2 x+\cos^2x}{\sin^2x}$$ $$(\cot x)'=-\frac{1}{\sin^2x}=-\csc^2x$$
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