Answer
$$y = \sqrt 3 x + \frac{1}{2}\pi x - \frac{{\sqrt 3 }}{2}$$
Work Step by Step
$$\eqalign{
& y = 3x\arcsin x,{\text{ }}\left( {\frac{1}{2},\frac{\pi }{4}} \right) \cr
& {\text{Differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {3x\arcsin x} \right] \cr
& \frac{{dy}}{{dx}} = 3x\frac{d}{{dx}}\left[ {\arcsin x} \right] + 3\arcsin x\frac{d}{{dx}}\left[ x \right] \cr
& \frac{{dy}}{{dx}} = 3x\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right) + 3\arcsin x \cr
& {\text{Calculate the slope at the given point }}\left( {\frac{1}{2},\frac{\pi }{4}} \right) \cr
& {\left. {m = \frac{{dy}}{{dx}}} \right|_{x = \frac{1}{2}}} = 3\left( {\frac{1}{2}} \right)\left( {\frac{1}{{\sqrt {1 - {{\left( {1/2} \right)}^2}} }}} \right) + 3\arcsin \left( {1/2} \right) \cr
& {\left. {m = \frac{{dy}}{{dx}}} \right|_{x = \frac{1}{2}}} = \sqrt 3 + \frac{1}{2}\pi \cr
& {\text{The equation of the tangent line at the point }}\left( {\frac{1}{2},\frac{\pi }{4}} \right){\text{ is}} \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y - \frac{\pi }{4} = \left( {\sqrt 3 + \frac{1}{2}\pi } \right)\left( {x - \frac{1}{2}} \right) \cr
& y - \frac{\pi }{4} = \sqrt 3 x + \frac{1}{2}\pi x - \frac{{\sqrt 3 }}{2} - \frac{1}{4}\pi \cr
& y = \sqrt 3 x + \frac{1}{2}\pi x - \frac{{\sqrt 3 }}{2} \cr} $$
