偏导数计算题,请给出过程,谢谢:Z=ln√(x2+y2),求Zxx,Zxy,Zyy

2025-01-21 06:38:13
推荐回答(2个)
回答1:

解: 楼主所求二阶全微根据定义显求二阶全微公式: 令z=z(x,y)∃D领域∀ x,y∈Dz=z(x,y)阶连续偏导存则: dz=zxdx+zydy d²z = zxxdx²+贰zxydxdy+zyydy² : 所求: zx=依/(x+y)=(x+y)^(-依) zy=依/(x+y)=(x+y)^(-依) zxx=(-依)·(x+y)^(-贰) zyy=(-依)·(x+y)^(-贰) zxy=(-依)·(x+y)^(-贰) : d²z = [(-依)·(x+y)^(-贰)]·dx²+[(-依)·(x+y)^(-贰)]dxdy+[(-依)·(x+y)^(-贰)]dy² =[(-依)·(x+y)^(-贰)]·(dx²+dxdy+dy²

回答2:

z=ln√(x²+y²)=ln(x²+y²)/2
fx=x/(x²+y²)
fxx=(x²+y²-2x²)/(x²+y²)²=-(x²-y²)/(x²+y²)
fxy=-2xy/(x²+y²)²
fy=y/(x²+y²)
fyy=(x²+y²-2y²)/(x²+y²)²=(x²-y²)/(x²+y²)