Answer
$$y = 2$$
Work Step by Step
$$\eqalign{
& {\text{function }}y = {2^{\sin x}}{\text{ at }}x = \pi /2 \cr
& {\text{evaluating }}y\left( {\pi /2} \right) \cr
& \,\,\,y\left( {\pi /2} \right) = {2^{\sin \left( {\pi /2} \right)}} \cr
& \,\,\,y\left( {\pi /2} \right) = {2^1} \cr
& \,\,\,y\left( {\pi /2} \right) = 2 \cr
& {\text{we have the point }}\left( {\pi /2,2} \right) \cr
& \cr
& {\text{find the slope at }}x = \pi /2 \cr
& \,\,\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{2^{\sin x}}} \right] \cr
& \,\,\frac{{dy}}{{dx}} = {2^{\sin x}}\left( {\ln 2} \right)\frac{d}{{dx}}\left[ {\sin x} \right] \cr
& \,\,\frac{{dy}}{{dx}} = {2^{\sin x}}\left( {\ln 2} \right)\cos x \cr
& \,\,m = {\left. {\frac{{dy}}{{dx}}} \right|_{x = \pi /2}} = {2^{\sin \left( {\pi /2} \right)}}\left( {\ln 2} \right)\cos \left( {\pi /2} \right) \cr
& \,\,m = {2^{\left( 1 \right)}}\left( {\ln 2} \right)\left( 0 \right) \cr
& \,\,m = 0 \cr
& {\text{find the equation of the tangent line at the point }}\left( {\pi /2,2} \right) \cr
& \,\,\,\,\,y - {y_1} = m\left( {x - {x_1}} \right) \cr
& \,\,\,\,\,y - 2 = 0\left( {x - \pi /2} \right) \cr
& \,\,\,\,\,y - 2 = 0 \cr
& \,\,\,\,\,y = 2 \cr
& \cr
& {\text{graph}} \cr} $$
