∂F
∂x =∂F 1
∂xˆe 1 +∂F^2
∂xˆe 2 +∂F^3
∂xˆe 3∂F
∂y =∂F 1
∂yˆe 1 +∂F^2
∂yˆe 2 +∂F^3
∂yˆe 3∂F
∂z =∂F 1
∂zˆe 1 +∂F^2
∂zˆe 2 +∂F^3
∂zˆe 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
∂xdx +∂
F
∂ydy +∂
F
∂zdzor
dF=(
∂F 1
∂xdx +∂F^1
∂ydy +∂F^1
∂zdz)
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 3calculate the partial derivatives
∂F
∂x ,∂F
∂y ,∂F
∂z ,∂^2 F
∂x∂ySolution: 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.