Use chain rule to split the problem into two:
[tex]\frac{d}{dx}\sin^{-1}(\ln x)=\frac{d sin^{-1}u}{du}\frac{du}{dx}[/tex]
with [tex]u = \ln x[/tex]
Let
[tex]y = \sin^{-1}u[/tex]
since sin^-1 is an inverse to sin we also know that
[tex]\sin y = u[/tex]
We are looking for
[tex]\frac{dy}{du} = \frac{d \sin^{-1}u}{du}\\\frac{1}{\frac{du}{dy}}= \frac{d \sin^{-1}u}{du}\\\frac{1}{\cos y}= \frac{d \sin^{-1}u}{du}\\\frac{1}{\sqrt{1-sin^2 y}}= \frac{d \sin^{-1}u}{du}\\\frac{1}{\sqrt{1-u^2}}= \frac{d \sin^{-1}u}{du}[/tex]
Now the second part of the chain rule:
[tex]\frac{du}{dx}=\frac{\ln x}{dx}=\frac{1}{|x|}=\frac{1}{x}\,\,\mbox{for}\,\,x>0[/tex]
and putting it all together
[tex]\frac{d sin^{-1}u}{du}\frac{du}{dx}= \frac{1}{x\sqrt{1-\ln^2 x}}[/tex]
the last form is the final derivative and your answer
