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: 44


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

Work Step by Step

Since $ y=\frac{\cos^{-1}x}{\sin^{-1}x}$, then by the qoutient rule, the derivative $ y'$ is given by $$ y'= \frac{\sin^{-1}x\left(\frac{-1}{\sqrt{1-x^2}}\right)-\cos^{-1}x\left(\frac{1}{\sqrt{1-x^2}}\right)}{(\sin^{-1}x)^2}\\=-\frac{\sin^{-1}x +\cos^{-1}x }{(\sin^{-1}x)^2 \sqrt{1-x^2}}.$$
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