Unizor - Partial Derivatives - Basic Properties

Notes to a video lecture on http://www.unizor.com

Partial Derivatives -
Basic Properties

We will mostly be concerned with partial derivatives of functions of two variable arguments.
The theory can be extended to functions of any number of arguments, but it's outside of the scope of this course.
Besides, functions of two arguments can be visualized as surfaces in three-dimensional space to better understand their properties.

Certain simple properties of partial derivatives of multivariable functions coincide with corresponding properties of derivatives of functions of one variable because the process of partial differentiation is, actually, a differentiation by one variable, keeping the others as constants.

So, we will not bother with proofs since they are based on corresponding properties of derivatives of functions of a single variable.
Also, whatever property is listed below for one argument of partial differentiation of multivariable function is valid for any other argument.

1. ∂[f(x,y)+g(x,y)]/∂x = ∂f(x,y)/∂x + ∂f(x,y)/∂x

Example 1
∂[ln(x²+y²)+x·y]/∂x = 2x/(x²+y²)+y

2. ∂[K·f(x,y)]/∂x = K·∂f(x,y)/∂x

Example 2
∂[2·ln(x²+y²)]/∂x = 2·2x/(x²+y²) = 4x/(x²+y²)

3. ∂[f(x,y)·g(x,y)]/∂x = g(x,y)·∂f(x,y)/∂x + f(x,y)·∂g(x,y)/∂x

Example 3
∂[(x−y)·(x²+xy+y²)]/∂x = (x−y)·(2x+y)+1·(x²+xy+y²) = 3x²
At the same time,
(x−y)·(x²+xy+y²) = x³−y³ and a partial derivative by x of this expression equals to 3x², which corresponds to a previous result.

4. ∂f(x(t),y)/∂t = ∂f(x,y)/∂x · dx(t)/dt

Example 4
∂[ln²(t)+y²]/∂t = 2ln(t)/t

Let's consider properties specific for multivariable functions.
Recall that for a function of a single variable f(x) the following approximation can be established:
Δf(x) = f(x+Δx) − f(x) ≅ (df(x)/dx)·Δx
As Δx→0, this approximation is transformed into a relationship between infinitesimals:
df(x) = (df(x)/dx)·dx

Similarly, we can establish a relationship in case of multivariable functions for each of its arguments:
f(x+Δx,y) − f(x,y) ≅ (∂f(x,y)/∂x)·Δx
and
f(x,y+Δy) − f(x,y) ≅ (∂f(x,y)/∂y)·Δy

Using the above properties, let's consider a general case when both arguments of function f(x,y) are incremented.
f(x+Δx,y+Δy) − f(x,y) =
= f(x+Δx,y+Δy) − f(x,y+Δy) + f(x,y+Δy) − f(x,y) ≅
≅ (∂f(x,y+Δy)/∂x)·Δx + (∂f(x,y)/∂y)·Δy

As Δx→0 and Δy→0, we can replace these increments with infinitesimal differentials and, using the smoothness of our functions (in particular, continuity of partial derivatives), we can write the following equivalence between infinitesimals:
df(x,y)=f(x+dx,y+dy)−f(x,y) =
= (∂f(x,y)/∂x)·dx + (∂f(x,y)/∂y)·dy
This expression is called total differential of a function of two variables.

Consider now that both arguments of our function f(x,y) are, in turn, functions of some argument:
x = x(t)
y = y(t)
We can use the same expression for a total differential
df(x,y) = (∂f(x,y)/∂x)·dx + (∂f(x,y)/∂y)·dy
but, considering that x and y are functions of t and, therefore,
dx = x'(t)·dt and
dy = y'(t)·dt,
we get the following:
df(x,y) = (∂f(x,y)/∂x)·x'(t)dt + (∂f(x,y)/∂y)·y'(t)dt

From the last equivalence we derive the formula for total derivative
df(x,y)/dt = (∂f(x,y)/∂x)·x'(t) + (∂f(x,y)/∂y)·y'(t)
This formula represents the chain rule for a function of two arguments when these arguments are, in turn, functions of some other same argument.

Example 5

Consider certain quantity of ideal gas in a reservoir with a piston, so we can change its volume, and a heater, so we can change its temperature.
The law of physics tells that the pressure P, volume V and absolute temperature T are connected by a formula
P·V/T = const
where the constant on the right depends on quantity of gas and in our case can be fixed and equal to C.

Let's apply some pressure to a piston to squeeze the gas.
The function that describes the change of volume with time is V(t).
Let's heat the gas.
The function that describes the change of temperature with time is T(t).
How fast the pressure would rise?

Since P(t)·V(t)/T(t)=C,
P(t) = C·T(t)/V(t)
The speed, with which the pressure is rising is a derivative dP(t)/dt
To calculate this derivative, we can use the formula of total derivative of a function of two arguments:
dP(t)/dt = (∂P(V,T)/∂V)·V'(t) + (∂P(V,T)/∂T)·T'(t) =
= −C·T(t)·V'(t)/V²(t) + C·T'(t)/V(t)