If ∫π/2x 3−2sin2zdz+∫0y costdt=0 then dydx at (π/2,π) is
1
2
3
0
Differentiating w ·r· t· x, we get
3−2sin2x+dydxcosy=0
Putting x=π/2 and y=π we get
dydx(π/2,π)=1.