Answer
$y'=\frac{-1}{t^2\sqrt (1-1/t^2)}$
Work Step by Step
Use the definition of inverse sine and apply the chain rule to find the derivative:
$y'=\frac{1}{\sqrt (1-(1/t)^2)}\times-1/t^2=\frac{-1}{t^2\sqrt (1-1/t^2)}$
Calculus: Early Transcendentals 8th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1285741552
ISBN 13:
978-1-28574-155-0
Chapter 3 - Section 3.5 - Implicit Differentiation - 3.5 Exercises - Page 216: 56
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