Answer
$h'\left( x \right) = \frac{{\arcsin x}}{x} + \frac{{\ln x}}{{\sqrt {1 - {x^2}} }}$
Work Step by Step
$$\eqalign{
& h\left( x \right) = \left( {\arcsin x} \right)\ln x \cr
& {\text{Differentiate}} \cr
& h'\left( x \right) = \frac{d}{{dx}}\left[ {\left( {\arcsin x} \right)\ln x} \right] \cr
& {\text{Apply the product rule }} \cr
& h'\left( x \right) = \left( {\arcsin x} \right)\frac{d}{{dx}}\left[ {\ln x} \right] + \ln x\frac{d}{{dx}}\left[ {\left( {\arcsin x} \right)} \right] \cr
& {\text{Compute derivatives}} \cr
& h'\left( x \right) = \left( {\arcsin x} \right)\left( {\frac{1}{x}} \right) + \ln x\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right) \cr
& {\text{Simplify}} \cr
& h'\left( x \right) = \frac{{\arcsin x}}{x} + \frac{{\ln x}}{{\sqrt {1 - {x^2}} }} \cr} $$
