高等数学偏导数问题

2025-05-18 00:50:33
推荐回答(2个)
回答(1):

  对方程两边求微分,得
   dz = f1*(ydx+xdy)+f2*[e^(xy)](ydx+xdy)
= y{f1+[e^(xy)]*f2}dx+x{f1+[e^(xy)]*f2}dy
可得
   Dz/Dx = y{f1+[e^(xy)]*f2},Dz/Dy = x{f1+[e^(xy)]*f2}。
于是
   D²z/Dx² = (D/Dx)[y{f1+[e^(xy)]*f2}]
       = y{(D/Dx)f1+(D/Dx)[e^(xy)]*f2+[e^(xy)]*(D/Dx)f2}
       = y{y{f11+[e^(xy)]*f12}+y[e^(xy)]*f2+[e^(xy)]*y{f21+[e^(xy)]*f22}}
       = ……。

回答(2):