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: 9

Answer

The Derivative is: $y'=\sec x(1+x\tan x)-\frac{1}{x^2}$

Work Step by Step

$y=x\sec x +\frac{1}{x}$ Applying Derivative Rules $y'=f'(x)+g'(x)$ and $f'(x)=h'(x)\cdot v(x) + h(x)\cdot v'(x)$ $y'=((x^{1-1})(\sec x)+(x)(\sec x\tan x))+(-1)x^{-1-1}$ $y'=(\sec x + x\sec x \tan x)-\frac{1}{x^2}$ $y'=\sec x(1+x\tan x)-\frac{1}{x^2}$
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