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: 4

Answer

The Derivative is: $y'=\frac{\sec x}{2\sqrt{x}}+\sqrt{x}\sec x\tan x$

Work Step by Step

$y=\sqrt{x}\sec x+3$ Applying Derivative Rules $y'=f'(x)+g'(x)$ and $f'(x)=h'(x)\cdot r(x)+h(x)\cdot r'(x)$ $y'=((\frac{1}{2\sqrt{x}})(\sec x)+(\sqrt{x})(\sec x\tan x))+(0)$ $y'=\frac{\sec x}{2\sqrt{x}}+\sqrt{x}\sec x\tan x$
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