129.Order of differentiation immaterial.
f
(
x
,
y
)
{\displaystyle f(x,y)}
x
{\displaystyle x}
x
+
δ
x
{\displaystyle x+\delta x}
y
{\displaystyle y}
(46 ), § 106 ,
(A )
f
(
x
+
δ
x
,
y
)
−
f
(
x
,
y
)
=
δ
x
f
x
′
(
x
+
θ
⋅
δ
x
,
y
)
.
{\displaystyle f\left(x+\delta x,y\right)-f\left(x,y\right)=\delta xf_{x}'\left(x+\theta \cdot \delta x,y\right).}
0
<
θ
<
1
{\displaystyle 0<\theta <1}
a
=
x
,
δ
a
=
δ
x
,
{\displaystyle a=x,\delta a=\delta x,}
x
{\displaystyle x}
y
{\displaystyle y}
x
{\displaystyle x}
If we now change
y
{\displaystyle y}
y
+
δ
y
{\displaystyle y+\delta y}
x
{\displaystyle x}
δ
x
{\displaystyle \delta x}
(A ) is
(B )
[
f
(
x
+
δ
x
,
y
+
δ
y
)
−
f
(
x
,
y
+
δ
y
)
]
−
[
f
(
x
+
δ
x
,
y
)
−
f
(
x
,
y
)
]
{\displaystyle \left[f\left(x+\delta x,y+\delta y\right)-f\left(x,y+\delta y\right)\right]-\left[f\left(x+\delta x,y\right)-f\left(x,y\right)\right]}
The total increment of the right-hand member of (A ) found by the Theorem of Mean Value, (46 ), § 106 , is
(C )
δ
x
f
x
′
(
x
+
θ
⋅
δ
x
,
y
+
δ
y
)
−
δ
x
f
x
′
(
x
+
θ
⋅
δ
x
,
y
)
{\displaystyle \delta xf_{x}'\left(x+\theta \cdot \delta x,y+\delta y\right)-\delta xf_{x}'\left(x+\theta \cdot \delta x,y\right)}
0
<
θ
1
<
1
{\displaystyle 0<\theta _{1}<1}
=
δ
y
δ
x
f
y
x
″
(
x
+
θ
1
⋅
δ
x
,
y
+
θ
2
⋅
δ
y
)
.
{\displaystyle =\delta y\delta xf_{yx}''\left(x+\theta _{1}\cdot \delta x,y+\theta _{2}\cdot \delta y\right).}
0
<
θ
2
<
1
{\displaystyle 0<\theta _{2}<1}
a
=
y
,
δ
a
=
δ
y
,
{\displaystyle a=y,\delta a=\delta y,}
y
{\displaystyle y}
x
{\displaystyle x}
δ
x
{\displaystyle \delta x}
y
{\displaystyle y}
Since the increments (B ) and (C ) must be equal,
(D )
[
f
(
x
+
δ
x
,
y
+
δ
y
)
−
f
(
x
,
y
+
δ
y
)
]
−
[
f
(
x
+
δ
x
,
y
)
−
f
(
x
,
y
)
]
{\displaystyle \left[f\left(x+\delta x,y+\delta y\right)-f\left(x,y+\delta y\right)\right]-\left[f\left(x+\delta x,y\right)-f\left(x,y\right)\right]}
=
δ
x
δ
y
f
y
x
″
(
x
+
θ
1
⋅
δ
x
,
y
+
θ
2
⋅
δ
y
)
.
{\displaystyle =\delta x\delta yf_{yx}''\left(x+\theta _{1}\cdot \delta x,y+\theta _{2}\cdot \delta y\right).}
In the same manner, if we take the increments in the reverse order,
(E )
[
f
(
x
+
δ
x
,
y
+
δ
y
)
−
f
(
x
+
δ
x
,
y
)
]
−
[
f
(
x
,
y
+
δ
y
)
−
f
(
x
,
y
)
]
{\displaystyle \left[f\left(x+\delta x,y+\delta y\right)-f\left(x+\delta x,y\right)\right]-\left[f\left(x,y+\delta y\right)-f\left(x,y\right)\right]}
=
δ
x
δ
y
f
x
y
″
(
x
+
θ
4
⋅
δ
x
,
y
+
θ
4
⋅
δ
y
)
.
{\displaystyle =\delta x\delta yf_{xy}''\left(x+\theta _{4}\cdot \delta x,y+\theta _{4}\cdot \delta y\right).}
θ
3
{\displaystyle \theta _{3}}
θ
4
{\displaystyle \theta _{4}}
The left-hand members of (D ) and (E ) being identical, we have
(F )
f
y
x
″
(
x
+
θ
1
⋅
δ
x
,
y
+
θ
2
⋅
δ
y
)
=
f
x
y
″
(
x
+
θ
3
⋅
δ
x
,
y
+
θ
4
⋅
δ
y
)
.
{\displaystyle f_{yx}''\left(x+\theta _{1}\cdot \delta x,y+\theta _{2}\cdot \delta y\right)=f_{xy}''\left(x+\theta _{3}\cdot \delta x,y+\theta _{4}\cdot \delta y\right).}
Taking the limit of both sides as
δ
x
{\displaystyle \delta x}
δ
y
{\displaystyle \delta y}
(G )
f
y
x
″
(
x
,
y
)
=
f
x
y
″
(
x
,
y
)
,
{\displaystyle f_{yx}''\left(x,y\right)=f_{xy}''\left(x,y\right),}
since these functions are assumed continuous. Placing
u
−
f
(
x
,
y
)
,
{\displaystyle u-f\left(x,y\right),}
(G ) may be written
(60 )
∂
2
u
∂
y
∂
x
=
∂
2
u
∂
x
∂
y
.
{\displaystyle {\frac {\partial ^{2}u}{\partial y\partial x}}={\frac {\partial ^{2}u}{\partial x\partial y}}.}
That is, the operations of differentiating with respect to
x
{\displaystyle x}
y
{\displaystyle y}
.
![{\displaystyle \left[f\left(x+\delta x,y+\delta y\right)-f\left(x,y+\delta y\right)\right]-\left[f\left(x+\delta x,y\right)-f\left(x,y\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bda4121d0821a07fb194d4ba36cf6ebc6c56aaf2)
![{\displaystyle \left[f\left(x+\delta x,y+\delta y\right)-f\left(x+\delta x,y\right)\right]-\left[f\left(x,y+\delta y\right)-f\left(x,y\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5e23b3b35dc4ac6ef7aeffb07d31efab03b820d3)
