## Calculus (3rd Edition)

$\frac{5}{4}.$
Since $y=\sin^{-1}x$, then $$y'= \frac{1}{\sqrt{1-x^2}}.$$ Now, we have $$y'(3/5)= \frac{1}{\sqrt{1-(3/5)^2}}=\frac{1}{(4/5)}=\frac{5}{4}.$$