When y=sin-1(x/a), sin(y)=x/a, cos(y)=√(1-x2/a2), cos(y)dy/dx=1/a, √(1-x2/a2)dy/dx=1/a, dy/dx=1/√(a2-x2).
But this problem has:
sin(y)=½(√(1+x)+√(1-x)),
sin2(y)=¼(1+x+1-x+2√(1-x2))=½(1+√(1-x2)),
cos(y)=√(1-sin2(y))=√(1-½(1+√(1-x2))=√(½(1-√(1-x2))),
cos(y)dy/dx=¼(1/√(1+x)-1/√(1-x))=¼(√(1-x)-√(1+x))/√(1-x2),
√(½(1-√(1-x2)))dy/dx=¼(√(1-x)-√(1+x))/√(1-x2),
dy/dx=¼(√(1-x)-√(1+x))/[√(1-x2)√(½(1-√(1-x2)))].