设y=y(x),z=z(x)是由方程z=xf(x+y)和F(x,y,z)=0所确定的函数,其中f和F分别具有一阶连续导数

2024-11-18 11:41:08
推荐回答(2个)
回答(1):

因为y=y(x),z=z(x)是由方程z=xf(x+y)和F(x,y,z)=0所确定的函数
等式z=xf(x+y)两边对x求导得:

dz
dx
=[xf(x+y)]'=f(x+y)+xf'(x+y)(x+y)'
=f(x+y)+xf'(野雹x+y)(1+
dy
dx

即:
dz
dx
=f(x+y)+xf'(x+y)(1+
dy
dx

等式F(x,y,z)=0两边对x求导得:
?F(x,y,z)
?x
+
?F(x,y,z)
?y
dy
dx
+
?F(x,y,手脊陆z)
?z
dz
dx
=0
根据等式:
?F(x,y,z)
?x
+
?F(x,y,z)
?y
dy
dx
+
?F(x,y,z)
?z
dz
dx
=0
以及等式:
dz
dx
=f(x+y)毕顷+xf'(x+y)(1+
dy
dx

可以解得:
dz
dx
=
[f(x+y)+xf′(x+y)]
?F(x,y,z)
?y
?xf′(x+y)
F(x,y,z)
?z
?F(x,y,z)
?y
+xf′(x+y)
F(x,y,z)
?z

回答(2):

简单计首肢算一下即可,答者首世芹历案如图所示