∂F
∂x =
∂F 1
∂x
ˆe 1 +∂F^2
∂x
ˆe 2 +∂F^3
∂x
ˆe 3
∂F
∂y =
∂F 1
∂y
ˆe 1 +∂F^2
∂y
ˆe 2 +∂F^3
∂y
ˆe 3
∂F
∂z =
∂F 1
∂z
ˆe 1 +∂F^2
∂z
ˆe 2 +∂F^3
∂z
ˆe 3.
(6 .81)
Observe that each component of the vector field F must be differentiated.
The higher partial derivatives are defined as derivatives of derivatives. For ex-
ample, the second order partial derivatives are given by the expressions
∂^2 F
∂x^2 =
∂
∂x
(
∂F
∂x
)
,
∂^2 F
∂y^2 =
∂
∂y
(
∂F
∂y
)
,
∂^2 F
∂x∂y =
∂
∂x
(
∂F
∂y
)
, (6 .82)
where each component of the vectors are differentiated. This is analogous to the
definitions of higher derivatives previously considered.
Total Derivative
The total differential of a vector field F =F(x, y, z)is given by
dF=∂
F
∂x
dx +∂
F
∂y
dy +∂
F
∂z
dz
or
dF=
(
∂F 1
∂x
dx +∂F^1
∂y
dy +∂F^1
∂z
dz
)
ˆe 1
+
(
∂F 2
∂x dx +
∂F 2
∂y dy +
∂F 2
∂z dz
)
ˆe 2
+
(
∂F 3
∂x dx +
∂F 3
∂y dy +
∂F 3
∂z dz
)
ˆe 3.
(6 .83)
Example 6-26.
For the vector field
F =F(x, y, z ) = (x^2 y−z)ˆe 1 + (yz^2 −x)ˆe 2 +xyz eˆ 3
calculate the partial derivatives
∂F
∂x ,
∂F
∂y ,
∂F
∂z ,
∂^2 F
∂x∂y
Solution: Using the above definitions produces the results
∂F
∂x = 2xy ˆe^1 −ˆe^2 +yz ˆe^3 ,
∂F
∂y =x
(^2) ˆe 1 +z (^2) ˆe 2 +xz ˆe 3 ,
∂F
∂z =−ˆe^1 + 2 yz ˆe^2 +xy ˆe^3
∂^2 F
∂x∂y =
∂^2 F
∂y∂x = 2x
ˆe 1 +zeˆ 3.