A side comment to start with: the notation sin^-1 for the inverse sine function (more explicitly, the inverse function of the restriction of sine to [-pi/2,pi/2]) is widespread but misleading. Indeed, the standard convention for exponents when using trig functions (e.g., sin^2 x:=(sin x)^2 suggests that sin^(-1) x is (sin x)^(-1)=1/(sin x). Of course, it is not, but the notation is very misleading. The alternative (and commonly used) notation arcsin x is much better.
Now for the derivative. This is a composite, so we will use the Chain Rule. We will need (ln x)'=1/x (see calculus of logarithms ) and (arcsin x)'=1/sqrt(1-x^2) (see calculus of inverse trig functions ).
Using the Chain Rule:
(ln(arcsin x))'=1/arcsin x \times (arcsin x)'=1/(arcsin x sqrt(1-x^2)).
