Calculus: Early Transcendentals 8th Edition

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

Chapter 1 - Section 1.5 - Inverse Functions and Logarithms - 1.5 Exercises: 70

Answer

$tan(sinx^{-1}) = \frac{1}{\sqrt {1-x^{2}}}$

Work Step by Step

If $y = sin^{-1}x$ then $siny = x$ and $\frac{-\pi}{2}\leq y \leq \frac{\pi}{2} $ We want to find $tan y$ but, since $siny$ is known, it is easier to find $sec y$ frst: $1^{2} = x^{2} + sec^{2}y$ $1^{2} - x^{2} = sec^{2}y$ $secy = \sqrt {1-x^{2}}$ Thus $tan(sinx^{-1}) = siny = \frac{1}{secy} = \frac{1}{\sqrt {1-x^{2}}}$
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