Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.10 Derivatives of Inverse Trigonometric Functions - 3.10 Execises: 1

Answer

$\frac{d}{dx}sin^{-1}x = \frac{1}{\sqrt {1-x^2}}$ $\frac{d}{dx}tan^{-1}x = \frac{1}{1+x^2}$ $\frac{d}{dx}sec^{-1}x =\frac{1}{|x|\sqrt {x^2-1}}$

Work Step by Step

The formulae for these $\text{derivatives}$ are listed under $\texttt{Theorem 3.22}$. There is no work involved in getting these derivatives, besides simply memorizing these formulae.
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