Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.6 - The Chain Rule - Exercises 3.6 - Page 149: 39

Answer

$\sqrt{x} sec^2(2\sqrt{x})+tan(2\sqrt{x})$

Work Step by Step

$ h(x)=xtan(2\sqrt{x})+7$ $ h'(x)=x\frac{d}{dx}(tan(2x^{1/2})+tan(2x^{1/2}).\frac{d}{dx}(x)+0)$ $=xsec^2(2x^{1/2})\frac{d}{dx}(2x^{1/2})+tan(2x^{1/2})$ =$ xsec^2(2\sqrt{x}).\frac{1}{\sqrt{x}}+tan(2\sqrt{x})$ =$\sqrt{x} \sec^2(2\sqrt{x})+tan(2\sqrt{x})$
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