Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 2 - The Derivative - 2.6 The Chain Rule - Exercises Set 2.6: 16

Answer

$f'(x)=\frac{\sec^2\sqrt x}{2\sqrt x}$

Work Step by Step

$f(x)=tan\sqrt x$ $f(x)=tan(x^{\frac{1}{2}})$ $f'(x)=sec^2(x^{\frac{1}{2}})\times\frac{d}{dx}(x^{\frac{1}{2}})$ $f'(x)=sec^2(x^{\frac{1}{2}})\times\frac{1}{2}x^{\frac{1}{2}-1}$ $f'(x)=sec^2(\sqrt x)\times\frac{1}{2}x^{\frac{-1}{2}}$ $f'(x)=sec^2\sqrt x)\times\frac{1}{2{x^\frac{1}{2}}}$ $f'(x)=sec^2(\sqrt x)\times\frac{1}{2\sqrt x}$ $f'(x)=\frac{sec^2\sqrt x}{2\sqrt x}$
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