Find the equation of the normal line to the graph of

y=x^2+x.siny+π\2
at x=0

y = x^2 + xsiny + pi/2...................differentiate implicitly

y' = 2x + xcosy(y') + siny

y'(1 - xcosy) = 2x + siny............y' = (2x + siny)/(1 - xcosy)

y(0) = pi/2.............normal goes through (0, pi/2)

y'(0) = (0 + 1)/(1 - 0) = 1.........tangent line has slope +1, normal has slope - 1.

Equation of a line with slope -1, through (0, pi/2) is y = - x + pi/2