Answer
$$y'=\frac{1}{ \sqrt{1- x^2}\sin^{-1}x}.$$
Work Step by Step
Since $ y=\ln(arcsin \ x)=\ln(\sin^{-1}x) $, then
$$ y'=\frac{1}{\sin^{-1}x}(\sin^{-1}x)'=\frac{1}{\sin^{-1}x}\frac{1}{ \sqrt{1- x^2}}
\\=\frac{1}{ \sqrt{1- x^2}\sin^{-1}x}.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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