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))#.
