Part 2 - Continuity & partial derivatives.

Introduction

In this part we’ll define continuity when dealing with functions with several variables. We’ll also start looking at derivatives of functions of several variables.

Continuity

Let’s recall what continuity of a function with one variable is defined:

A function of one variable is continuous at a point, $a \in D$ if:

$$ \lim_{x \to a} f(x) = f(a) $$

In functions with several variables, it’s the same: $$ \lim_{(x, y) \to (a, b)} f(x, y) = f(a, b) $$

We say $f$ is continuous on its domain, if $f$ is continuous at each point, $(a, b) \in D$

Continuity laws

Suppose $f and g$ are functions of two variables. $h$, is a function of one variable. Which are continuous in their respective domains, then:

$$ f \pm g $$

$$ f \cdot g $$

$$ \left(\dfrac{f}{g}\right) $$

$$ h(f(x, y)) $$

All of the above, are continuous on their domains.

Example

$$ f(x, y) = x \ | \ \text{ is continuous at $\mathbb{R}^2$} $$

Proof: $$ \lim_{(x, y) \to (a, b)} f(x, y) = f(a, b) = a $$

$$ \text{Using the limit law: } \lim_{(x, y) \to (a, b)} x = a $$

Similarly $f(x, y) = y$ is continuous at $\mathbb{R}^2$.

$$ f(x, y) = x^3y - y^2 + x \ | \ \text{ is continuous at $\mathbb{R}^2$} $$

Proof: Since polynomials are combinations of $f(x, y) = x$ and $f(x, y) = y$, using continuity laws (or limit laws), we can easily prove this.

$$ f(x, y) = \dfrac{x^3 y^3 - x^4y}{y - x} \ | \ \text{ is continuous \textbf{on its domain}} $$

The domain of $f$ is: $$ \{(x, y) \ | \ y - x \neq 0 \} $$

$$ f(x, y) = e^{x^2 - y} \ | \ \text{ is continuous at $\mathbb{R}^2$} $$

Using the composition law of continuity, we easily prove that this function is a continuous function, on its domain. In this case it’s still $\mathbb{R}^2$.

Knowing whether a function is continuous or not is very powerful, knowing that the above function is continuous makes:

$$ \lim_{(x, y) \to (a, b)} e^{x^2 - y} = e^{1^2 - 3} = \boxed{e^{-2}} $$

Trivial.

Determine if this function is continuous: $$ f(x, y) = \begin{cases} \dfrac{xy^2}{x^2 + 3y^4} & (x, y) \neq (0,0) \newline 0 & (x, y) = (0, 0) \end{cases} $$

Solution: If the function is continuous, using the definition, then: $$ \lim_{(x, y) \to (a, b)} f(x, y) = f(a, b) $$

which means: $$ \lim_{(x, y) \to (0, 0)} f(x, y) = f(0, 0) = 0 $$

$$ \lim_{(x, y) \to (0, 0)} \dfrac{xy^2}{x^2 + 3y^4} = 0 $$

Using the law of limits and the definition, we find that this limit does not exist since:

Along $x = 0$ $$ \dfrac{xy^2}{x^2 + 3y^4} = \dfrac{0}{3y^4} = 0 $$

Along $x = y^2$ $$ \dfrac{xy^2}{x^2 + 3y^4} = \dfrac{y^4}{y^4 + 3y^4} = \dfrac{1}{4} $$

Which tells us that $f(x, y)$ is not continuous at $(0, 0)$.

Last example: $$ f(x, y) = \begin{cases} \dfrac{x^4 - y^4}{x^2 + y^2} & (x, y) \neq (0,0) \newline 0 & (x, y) = (0, 0) \end{cases} $$

$$ \lim_{(x, y) \to (0, 0)} \dfrac{x^4 - y^4}{x^2 + y^2} = 0 $$

$$ \begin{align*} \lim_{(x, y) \to (0, 0)} \dfrac{x^4 - y^4}{x^2 + y^2} & = \newline \lim_{(x, y) \to (0, 0)} \dfrac{(x^2 + y^2)(x^2 - y^2)}{x^2 + y^2} & = \newline \lim_{(x, y) \to (0, 0)} (x^2 - y^2) & = \newline \boxed{0} \end{align*} $$

Which means it is continuous at $(0, 0)$.

Partial derivatives

Let’s first start with recalling the definition of derivative for functions with one variable: $$ \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h} $$

Since we’re dealing with several variables, we need to choose what variable to respect, meaning what variable has a small increase.

Definition

The partial derivate of a function of two variables with respect to x, denoted by $f_x$, is the function of two variables given by: $$ f_x(x, y) = \lim_{h \to 0} \dfrac{f(x + h, y) - f(x, y)}{h} $$

Similarly for $y$: $$ f_y(x, y) = \lim_{h \to 0} \dfrac{f(x, y + h) - f(x, y)}{h} $$

In other words what this means, the partial derivate with respect to a variable, we treat all other variables as they were constants.

Examples

$$ f(x, y) = x^2y + y^3 $$

$$ f_x(x, y) = 2xy $$

$$ f_y(x, y) = x^2 + 3y^2 $$

$$ f(x, y) = sin(x^2y - 3x) $$

$$ f_x(x, y) = cos(x^2y - 3x) \cdot (2xy - 3) \ | \ \text{chain rule} $$

$$ f_y(x, y) = cos(x^2y - 3x) \cdot x^2 \ | \ \text{chain rule} $$

$$ f(x, y, z) = x^2z + xy - 7yz^5 $$

$$ f_z(x, y, z) = x^2 - 35yz^4 $$

Summary

As we have seen continuity is quite similar when dealing with functions of several variables. The partial derivate of functions of several variables are also quite trivial since we treat the other variables as constants!