Calculus I:
functions of several variables
Partial derivatives

Partial derivatives

The idea is to come back to a known situation.

We will proceed this way. We will define the notion of partial derivative by using the case of single-variable functions.


Exercise 1

f: R2 → R
(x, y)→ x2 + y5 + xy + 5

1. Determine f_{x = 1} (y).
2. Calculate f'_{x = 1} (y) and f'_{x = 1}(2).
3. General case:
Determine f_{x = xo}(y).
Calculate f'_{x = xo}(y) and f'_{x = xo}(2).
4. Determine f_{y = 1} (x).
5. Calculate f'_{y = 1}(x) and f'_{y = 1}(2)
6. General case:
Determine f_{y = yo}(x).
Calculate f'_{y = yo}(2)

Solution:

1. f_{x = 1} (y)
2. f'_{x = 1} (y) = 5y4 + 1
f'_{x = 1}(2) = 5 (2)4 +1 = 81.
3. General case:
f_{x = xo}(y)= xo2 + y5 + xoy + 5
f'_{x = xo}(y) = 5y4 + xo
f'_{x = xo}(2). = 5(2)4 + xo = 80 + xo
4. f_{y = 1} (x) = x2 + x + 6
5. f'_{y = 1}(x) = 2x + 1
f'_{y = 1}(2) = 2 x 2 + 1 = 5
6. General case:
f_{y = yo}(x) = x2 + yo5 + xyo + 5
f'_{y = yo} (2) = 2 x 2 + yo = 4x + yo


• Definition 1

f: D → R (x, y)→ f (x, y)

The Partial derivative of f with respect to x at the point (a, b) is the derivative f'_y=b (a).

This partial derivative is noted: ∂f/∂x(a,b) or ∂_xf(a,b).

The partial derivative of f with respect to y at point (a, b) is the derivative f'_x=a (b)

This partial derivative is noted: ∂f/∂y(a,b) or ∂_yf(a,b).

Note:

f'_y=b (a) means that y is constant and is equal to b, so we take the derivative with respect to the remaining variable that is x. This notation highlights the fact that y remains constant.

The notation ∂f/∂x(a,b) highlights the fact that we take the derivative with respect to x.


• Definition 2

We write ∂f/∂x the function that for a couple (x, y) associates the number ∂f/∂x(x,y) .

Similarly, we note ∂f/∂y the function that has a couple (x, y) associates the number ∂f/∂y(x,y).

With the the notion of partial derivative exit we can speak of a derivative for the functions of two variables. For functions of three or more variables the the mechanism is exactly the same.

For the functions of a variable we can easily calculate a second derivative: Just take the derivative of the function f'. What happens with the functions of two variables ?


• Definition 3

If f (x, y) admits partial derivatives at any point (x, y) of a domain, then ∂f/∂x and ∂f/∂y are themselves functions of x and y. So ∂f/∂x and ∂f/∂y can therefore also have partial derivatives.

These seconds are noted:

∂/∂x (∂f/∂x) = ∂²f/∂ x²
∂/∂y (∂f/∂y) = ∂²f/∂ y²
∂/∂x (∂f/∂y) = ∂²f/∂x∂y
∂/∂y (∂f/∂x) = ∂²f/∂y∂x


Exercice 2

Let f : R² → R

(x, y) → x² + y⁵ + xy + 5

1. Calculate ∂f/∂x.
2. Calculate ∂f/∂y.
3. Calculate ∂²f/∂ x².
4. Calculate ∂²f/∂ y².
5. Calculate ∂²f/∂x∂y.
6. Calculate ∂²f/∂y∂x.


Solution :

1. ∂f/∂x = 2x + y.
2. ∂f/∂y = 5y4 + x.
3. ∂²f/∂ x² = ∂f/∂ x (2x + 4) = 2
4. ∂²f/∂ y² = 20y3.
5. ∂²f/∂x∂y = 1.
6. ∂²f/∂y∂x = 1.

We remark that ∂²f/∂x∂y = ∂²f/∂y∂x.


Theorem 1

If ∂²f/∂x∂y and ∂²f/∂y∂x are continuous then ∂²f/∂x∂y = ∂²f/∂y∂x.

In other words, in this case, the order of taking derivatives is not important.

To understand this theorem we need to define the continuity of a function:


• Definition 4

Let f be a function of two variables, we say that f is continuous in (x_o, y_o) when the following condition is verified:

lim f (x, y) = f (x_o, y_o).
(x, y) → (x_o, y_o)

This definition means that no matter how close we get to (x_o, y_o) we have to get the same limit value which is f (x_o, y_o).