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: 10

Answer

The Derivative is: $y'=\sec^2 x$

Work Step by Step

$y=(\sin x +\cos x)\sec x$ Applying the distributive property of mathematics $y=\sin x \sec x + \cos x\sec x$ $y=\sin x\frac{1}{\cos x}+\cos x\frac{1}{\cos x}$ $y=\tan x +1$ Applying Derivative Rules: $y'=f'(x) + g'(x)$ $y'=\frac{d}{dx}(\tan x) + \frac{d}{dx}(1)$ $y'=(sec^2 x) + 0$ $y'=sec^2 x$
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