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: 14

Answer

The Derivative is: $y'=\frac{-x\sin x-\cos x}{x^2}+\sec x+x\sec x\tan x$

Work Step by Step

$y=\frac{\cos x}{x}+\frac{x}{\cos x}$ $y'=\frac{(-\sin x)(x)-(\cos x)(x^{1-1})}{(x)^2}+\frac{(x^{1-1})(\cos x)-(x)(-\sin x)}{\cos^2 x}$ $y'=\frac{-x\sin x-\cos x}{x^2}+\frac{\cos x+x\sin x}{\cos^2 x}$ $y'=\frac{-x\sin x-\cos x}{x^2}+\frac{\cos x}{\cos^2 x}+\frac{x\sin x}{\cos^2 x}$ $y'=\frac{-x\sin x-\cos x}{x^2}+\frac{\cos x}{\cos x}\frac{1}{\cos x}+\frac{1}{\cos x}\frac{\sin x}{\cos x}x$ Applying trigonometric identities: $y'=\frac{-x\sin x-\cos x}{x^2}+\sec x+x\sec x\tan x$
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