Explanation, Formula, Equation, Proof - Partial derivatives: Total derivatives
TOTAL DERIVATIVES
if u = f(x,y) where x = ϕ(t) and y = ψ(t) then we can express u
as a function of t alone by substituting the values of x and y in f(x, y). Thus we can find the ordinary
derivative du/dt which is called the total derivative of u to distinguish it from the partial derivatives ∂u/∂x
and ∂u/∂y.
Now to find du/dt without actually substituting the
values of x and y in f(x, y) we
establish the following Chain rule:
du/dt = ∂u/∂x.dx/dt + ∂u/∂x.dy/dt ... (i)

Proof.
We
have u = f(x,y)
Giving increment δt to t,
let the corresponding increments of x,y and u
be δx, δy and δu respectively. Then
u
+ δu = f(x + δx, y + δy)
Subtracting, δu = f(x
+ δx, y +δy) − f(x, y)
= | f(x + δx,
y+δy) − f(x,y+δy)| + |f(x,y+δy) ‒f(x,y) |

[Supposing ∂f(x,y)/∂x to be a continuous function of y]
= [ ∂f(x,y)/∂x . dx/dt ] + [ ∂f(x,y)/∂y . dy/dt
] which is the desired formula.
Cor.
Taking
t=x, (i) becomes du/dx = ∂u/∂x + ∂u/∂y.dy/dx
Obs. If u = f(x, y, z), where x. y, z are all functions of a variables t, then chain rule is

Let u= f(x,
y, ...) be a continuous function of several variables x, y, .. with continuous partial derivatives ∂u/∂x, ∂u/∂y,
…. If each variable is a function t,
that is, x = x(t), y = y(t), and so on.
Then the total
derivative of u with respect to t is given by

This is known as Chain
Rule for total derivatives.
If u = f(x,y,z) be such that y and z are function of x. Then ƒ is a function of one independent
variable. Then the total derivative of ƒ is given by

Composite
functions
If u = f(x, y), where x=ϕ(t), y=ψ(t), then u is called
a composite function of single variable t and we can find du/dt.
If u = f(x,y) where x = ϕ(r,s), y = ψ(r,s), then u is
called a composite function of two variables u and v so that we can find du/dr
and du/ds.
(1) If u is composite function of t, defined by the relations
u = f(x,y); x = ϕ(t), ψ(t),
then
……(1)
The differential form
(1) can be written as du = du/dx
dx + du/dy dy, du is called the total
differential of u.
If u = f(x, y, z) and x, y, z are functions of t,
then

2) If u = f(x,y),
where y = ϕ(x), then u is a composite
function of x alone.

3) If we are given an
implicit function f(x, y) = c, then u = f(x,
y), where u = c.
Since, u = c, du/dx = 0
From (a), we get

By the similar way, we
can prove

If u = f(r,s,t) and r, s, t
are functions of x, y and z, then

Applied Calculus: UNIT II: Functions of Several Variables : Tag: Applied Calculus : - Partial derivatives: Total derivatives
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