Calculus I:
functions of several variables
Differential

1. Differential of a function

In the first section of this chapter we have seen the differential notation.

For a function of a variable f (x) we have: f'(x) = df/dx or

df = f'(x) dx

Here again we will generalize what has been done to one variable:


• Definition 1

Let f be a function of two variables (x, y). We then note the differential of f in the following way:

df = (∂f/∂x) dx + (∂f/∂y) dy

Obviously if f is a function of three variables (x, y, z) then we have:

df = (∂f/∂x) dx + (∂f/∂y) dy + (∂f/∂z) dz


Exercise 1

Calculate the differencial of f (x, y, z) = x2y3z7 + x + sin (z) + 7.

Solution:

∂f/∂x = 2xy3z7 + 1.
∂f/∂y = 3x2y2z7.
∂f/∂z = 7x2y3z6 + cos (z)

Therefore:

df = (2xy3z7 + 1)dx + (3x2y2z7) dy + (7x2y3z6 + cos (z) )dz

2. Using a Differential
differential a compound function

Let's see on an example how to compute the differential of a compound function.

The approach is very simple, just substitute. . .


Exercise 2

1. Let f (u, v) = sin (u.v).
Calculate df.
2. u, and v now indicate functions. We write:
u (x, y) = x - 7y, and v (x, y) = x + y.
Calculate du, and dv.
3. The function is now considered:
F (x, y) = sin ((x - 7y). (x + y))
Calculate dF, then reduce ∂F/∂x

Solution:

1. df = (∂f/∂u) du + (∂f/∂v)dv

∂f/∂u = v. cos (uv),
∂f/∂v = u. cos (uv),

Therefore :

df = v. cos (u.v) du + u. cos (uv) dv.

2. We have

du = (∂u/∂x) dx + (∂u/∂y)dy
and
dv = (∂v/∂x) dx + (∂v/∂y)dy

∂u/∂x = 1,
∂u/∂y = - 7.

∂v/∂x = 1,
∂v/∂y = 1.

It comes then:

du = dx - 7dy.

Similarly we get:

dv = dx + dy.

3. We have:

∂f/∂u = v. cos (uv),
∂f/∂v = u. cos (uv),

So

dF = (∂F/∂u)du + ∂F/∂v)dv =

[(x + y). cos ((x - 7y)(x + y))](dx - 7dy) +
[(x - 7y). cos ((x - 7y)(x + y))]( dx + dy) =

[(2x - 6y). cos ((x - 7y)(x + y))]dx +
[(-6x - 14y). cos ((x - 7y)(x + y))] dy

dF = 2[(x - 3y). cos ((x - 7y)(x + y))]dx
- 2[(3x + 7y). cos ((x - 7y)(x + y))] dy

Since dF = (∂F/∂x)dx + (∂F/∂y)dy, by identifying we have :

∂F/∂x = 2(x - 3y). cos ((x - 7y)(x + y))


Conclusion:

F was the compound of the function f with the functions u and v. So we have calculate the differential dF , using the df, du, and dv differentials using the substitution principle.

Here, this method allowed us:

• to use simple calculations already done,
• to obtain, without making new calculations of derivatives, the value of ∂F/∂x.