Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Review - Exercises: 38

Answer

$y'=\dfrac{1}{2(1+\arcsin^ {2}\sqrt{x})\sqrt{x-x^{2}}}$

Work Step by Step

$y=\arctan(\arcsin\sqrt{x})$ Start the differentiation process by using the chain rule: $y'=\dfrac{1}{1+(\arcsin\sqrt{x})^{2}}(\arcsin\sqrt{x})'=\dfrac{(\arcsin\sqrt{x})'}{1+\arcsin^{2}\sqrt{x}}=...$ Apply the chain rule one more time to evaluate the indicated derivative: $...=\dfrac{\Big[\dfrac{1}{\sqrt{1-(\sqrt{x})^{2}}}\Big](\sqrt{x})'}{1+\arcsin^{2}\sqrt{x}}=\dfrac{\Big(\dfrac{1}{\sqrt{1-x}}\Big)(x^{1/2})}{1+\arcsin^{2}\sqrt{x}}=...$ Evaluate the derivative indicated and simplify: $...=\dfrac{\Big(\dfrac{1}{\sqrt{1-x}}\Big)\Big(\dfrac{1}{2}\Big)x^{-1/2}}{1+\arcsin^{2}\sqrt{x}}=\dfrac{1}{2(\sqrt{x})(\sqrt{1-x})(1+\arcsin^{2}\sqrt{x})}=...$ $...=\dfrac{1}{2(1+\arcsin^ {2}\sqrt{x})\sqrt{x-x^{2}}}$
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