关于多元函数微分的问题设x=(e^u)*cosv,y=(e^u)*sinv,z=uv,试求dz/dx和dz/dy(偏微分

∵x=(e^u)*cosv,y=(e^u)*sinv,
∴U=ln(x²+y²),V=arctan(y/x).
∴Ux=2x/(x²+y²),Vx=-y/(x²+y²),
Uy=2y/(x²+y²),Vy=x/(x²+y²).
∵Z=UV,
∴Zu=V,Zv=u.(Ux,Vx,Uy,Vy,Zu,Zv分别表示它们关于下标的偏导数).
故dz/dx=Zu*Ux+Zv*Vx
=V*(2x/(x²+y²))+U*(-y/(x²+y²))
=[2x*arctan(y/x)-y*ln(x²+y²)]/(x²+y²);
dz/dy=Zu*Uy+Zv*Vy
=V*(2y/(x²+y²))+U*(x/(x²+y²))
=[2y*arctan(y/x)+x*ln(x²+y²)]/(x²+y²).

详见：

1年前