Begin2.DVI

Differentiation of Composite Functions

Let φ=φ(x, y, z)define a scalar field and consider a curve passing through the

region where the scalar field is defined. Express the curve through the scalar field

in the parametric form

x=x(t), y =y(t), z =z(t),

with parameter t. The value of the scalar φ, at the points (x, y, z )along the curve, is

a function of the coordinates on the curve. By substituting into φthe position of a

general point on the curve, one can write

φ=φ(x(t), y(t), z(t)).

By substituting the time-varying coordinates of the curve into the function φ, one

creates a composite function. The time rate of change of this composite function φ,

as one moves along the curve, is derived from chain rule differentiation and

dφ

dt

=∂φ

∂x

dx

dt

+∂φ

∂y

dy

dt

+∂φ

∂z

dz

dt

. (6 .85)

The equation (6.85) gives us the general rule

d[ ]

dt =

∂[ ]

∂x

dx

dt +

∂[ ]

∂y

dy

dt +

∂[ ]

∂z

dz

dt (6 .86)

where the quantity inside the brackets can be any scalar function of x, y and z. The

second derivative of φcan be calculated by using the product rule and

d^2 φ

dt^2 =

∂φ

∂x

d^2 x

dt^2 +

dx

dt

d

dt

[

∂φ

∂x

]

+

∂φ

∂y

d^2 y

dt^2 +

dy

dt

d

dt

[∂φ

∂y

]

+∂φ

∂z

d^2 z

dt^2

+dz

dt

d

dt

[

∂φ

∂z

]

.

(6 .87)

To evaluate the derivatives of the terms inside the brackets of equation (6.87) use

the general differentiation rule given by equation (6.86). This produces a second

derivative having the form

d^2 φ

dt^2

= ∂φ

∂x

d^2 x

dt^2

+dx

dt

[

∂^2 φ

∂x^2

dx

dt

+ ∂

(^2) φ
∂x ∂y
dy
dt

• ∂
  (^2) φ
  ∂x ∂z
  dz
  dt
  ]


• 
  

  ∂φ
  ∂y
  d^2 y
  dt^2 +
  dy
  dt
  [
  ∂^2 φ
  ∂y ∂x
  dx
  dt +
  ∂^2 φ
  ∂y^2
  dy
  dt +
  ∂^2 φ
  ∂y ∂z
  dz
  dt
  ]

  
  


• 
  

  ∂φ
  ∂z
  d^2 z
  dt^2 +
  dz
  dt
  [
  ∂^2 φ
  ∂z ∂x
  dx
  dt +
  ∂^2 φ
  ∂z ∂y
  dy
  dt +
  ∂^2 φ
  ∂z^2
  dz
  dt
  ]
  .
  (6 .88)