University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 158: 39


$$h'(x)=\tan(2\sqrt x)+\sqrt x\sec^2(2\sqrt x)$$

Work Step by Step

$$h(x)=x\tan(2\sqrt x)+7$$ The derivative of $h(x)$ is: $$h'(x)=(x')\tan(2\sqrt x)+x\Big(\tan(2\sqrt x)\Big)'+0$$ $$h'(x)=\tan(2\sqrt x)+x\sec^2(2\sqrt x)(2\sqrt x)'$$ $$h'(x)=\tan(2\sqrt x)+x\sec^2(2\sqrt x)(2x^{1/2})$$ $$h'(x)=\tan(2\sqrt x)+x\sec^2(2\sqrt x)\Big(2\times\frac{1}{2}x^{-1/2}\Big)$$ $$h'(x)=\tan(2\sqrt x)+x\sec^2(2\sqrt x)\Big(\frac{1}{\sqrt x}\Big)$$ $$h'(x)=\tan(2\sqrt x)+\sqrt x\sec^2(2\sqrt x)$$
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