ysinx-cos(x+y)=0,两边对x求导,得
y'sinx+ycosx+(1+y')sin(x+y)=0,解得
y'=-[ycosx+sin(x+y)]/[sinx+sin(x+y)]
dy/dx=y'(0,π/2)=-(π/2+1)/(0+1)=-(π/2+1)