Question

请问这道高等数学题怎么求

Answer 1

20.z=y/√(x^2+y^2),
所以dz=dy/√(x^2+y^2)+yd[1/√(x^2+y^2)]
=dy/√(x^2+y^2)+y*(-1/2)(x^2+y^2)^(-3/2)*(2xdx+2ydy)
=(x^2+y^2)^(-3/2)*(-xydx+x^2dy).

Answer 2

dz＝(偏z／偏x)dx＋(偏z／偏y)dy
＝－x／(x^2+y^2)^(3/2)(ydx－xdy)

追答

Answer 3

方法如下，
请作参考：

Answer 4

z = y(x^2+y^2)^(-1/2)
∂z/∂x = (-1/2)y(x^2+y^2)^(-3/2) (2x) =-xy/(x^2+y^2)^(3/2)
∂z/∂y = (x^2+y^2)^(-1/2) + (-1/2)y(x^2+y^2)^(-3/2) (2y)
= (x^2+y^2)^(-1/2) - y^2(x^2+y^2)^(-3/2)
= x^2/(x^2+y^2)^(-3/2)
dz = (∂z/∂x)dx + (∂z/∂y)dy = [x/(x^2+y^2)^(3/2)](-ydx+xdy)