Answer
$$f'\left( x \right) = \frac{1}{{x\sqrt {1 - {{\ln }^2}x} }}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\sin ^{ - 1}}\left( {\ln x} \right) \cr
& {\text{find the derivative}}{\text{, differentiate both sides with respect to }}x \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{{\sin }^{ - 1}}\left( {\ln x} \right)} \right] \cr
& {\text{use }}\frac{d}{{dx}}\left[ {{{\sin }^{ - 1}}u} \right] = \frac{1}{{\sqrt {1 - {u^2}} }}\frac{{du}}{{dx}}.{\text{ let }}u = \ln x \cr
& f'\left( x \right) = \left( {\frac{1}{{\sqrt {1 - {{\left( {\ln x} \right)}^2}} }}} \right)\frac{d}{{dx}}\left[ {\ln x} \right] \cr
& f'\left( x \right) = \left( {\frac{1}{{\sqrt {1 - {{\left( {\ln x} \right)}^2}} }}} \right)\left( {\frac{1}{x}} \right) \cr
& {\text{simplifying}} \cr
& f'\left( x \right) = \frac{1}{{x\sqrt {1 - {{\ln }^2}x} }} \cr} $$
