Question

求z=（ax+by/ax-by）^1/2的全微分

Answer 1

dz=1/2[(ax-by)/(ax+by)]^1/2 {[ a(ax-by)-(ax+by)a ]dx/(ax-by)^2+[b(ax-by)-(ax+by)(-b)]dy/(ax-by)^2}

=ab(x dy-y dx)/[(ax+by)(ax-by)^3]^1/2