Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 374: 48


$$y'=\frac{1}{ \sqrt{1- x^2}\sin^{-1}x}.$$

Work Step by Step

Since $ y=\ln(arcsin \ x)=\ln(\sin^{-1}x) $, then $$ y'=\frac{1}{\sin^{-1}x}(\sin^{-1}x)'=\frac{1}{\sin^{-1}x}\frac{1}{ \sqrt{1- x^2}} \\=\frac{1}{ \sqrt{1- x^2}\sin^{-1}x}.$$
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