Part 2 - Continuity & partial derivatives.

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 $(0, 0)$.

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

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

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

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

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

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

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

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

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