原式=[-x(b/a)]*[-a/{x[(ax)^(1/2)]}*[-2ab(x/b)^(1/2)].(x>0)
=(-2)*[x^[1-1-(1/2)+(1/2)]*a^[-1+1-(1/2)+1]*b^[1+1-(1/2)].
=(-2)x^0*a^(1/2)*b^(3/2).
故,原式=(-2)a^(1/2)*b^(3/2).