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 - Page 141: 13

Answer

The Derivative is: $y'=4\sec x\tan x-\csc^2 x$

Work Step by Step

$y=\frac{4}{\cos x}+\frac{1}{\tan x}$ $y=4\frac{1}{\cos x}+\frac{1}{\tan x}$ $y=4\sec x+\cot x$ Applying Derivative rules: $y'=4\frac{d}{dx}(\sec x)+\frac{d}{dx}(\cot x)$ $y'=4(\sec x\tan x)+(-\csc^2x)$ $y'=4\sec x\tan x-\csc^2 x$
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