Answer
$$\,\,\,\,\,y = \frac{1}{{2\sqrt 3 }}x - \frac{1}{{\sqrt 3 }} + \frac{\pi }{6}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\sin ^{ - 1}}\left( {\frac{x}{4}} \right);{\text{ }}\left( {2,\frac{\pi }{6}} \right) \cr
& {\text{differentiate }}f\left( x \right) \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{{\sin }^{ - 1}}\left( {\frac{x}{4}} \right)} \right] \cr
& {\text{use }}\frac{d}{{du}}\left[ {{{\sin }^{ - 1}}u} \right] = \frac{1}{{\sqrt {1 - {u^2}} }}\frac{{du}}{{dx}} \cr
& f'\left( x \right) = \frac{1}{{\sqrt {1 - {{\left( {x/4} \right)}^2}} }}\frac{d}{{dx}}\left[ {\frac{x}{4}} \right] \cr
& f'\left( x \right) = \frac{1}{{\sqrt {\frac{{16 - {x^2}}}{{16}}} }}\left( {\frac{1}{4}} \right) \cr
& f'\left( x \right) = \frac{4}{{\sqrt {16 - {x^2}} }}\left( {\frac{1}{4}} \right) \cr
& f'\left( x \right) = \frac{1}{{\sqrt {16 - {x^2}} }} \cr
& \cr
& {\text{find the slope at the point }}\left( {2,\frac{\pi }{6}} \right) \cr
& m = f'\left( 2 \right) = \frac{1}{{\sqrt {16 - {{\left( 2 \right)}^2}} }} \cr
& \,\,m = \frac{1}{{\sqrt {12} }} \cr
& \,{\text{find the equation of the tangent line at the point }}\left( {2,\frac{\pi }{6}} \right) \cr
& \,\,\,\,\,y - {y_1} = m\left( {x - {x_1}} \right) \cr
& \,\,\,\,\,y - \frac{\pi }{6} = \frac{1}{{\sqrt {12} }}\left( {x - 2} \right) \cr
& \,\,\,\,\,y - \frac{\pi }{6} = \frac{1}{{\sqrt {12} }}x - \frac{2}{{\sqrt {12} }} \cr
& \,\,\,\,\,y = \frac{1}{{2\sqrt 3 }}x - \frac{2}{{2\sqrt 3 }} + \frac{\pi }{6} \cr
& \,\,\,\,\,y = \frac{1}{{2\sqrt 3 }}x - \frac{1}{{\sqrt 3 }} + \frac{\pi }{6} \cr} $$
