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.
