## Continuity for Functions of Several Variables, Partial Derivatives

1. We considered differentiating and integrating functions of a single-variable. How? We began with the notion of a limit, and then considered the derivative. If we have a, e.g., polynomial
(1)$\displaystyle p(x,y) = x^{3}+x^{2}y+xy^{2}+y^{3}$
we see
(2)$\displaystyle p(x+\Delta x,y) = x^{3}+x^{2}y+xy^{2}+y^{3}+\bigl(3x^{2}+2xy+y\bigr)\Delta x+ \mathcal{O}(\Delta x^{2})$
Again we stop and reflect: this treats ${y}$ as if it were constant. So the derivative formed by
(3)$\displaystyle \lim_{\Delta x\rightarrow0}\frac{p(x+\Delta x,y)-p(x,y)}{\Delta x}=3x^{2}+2xy+y$
are “incomplete” or partial. There is some subtlety here due to using multiple variables, and we have to discuss the problems of limits first.

2. Definition. The function ${z=f(x,y)}$ is “Continuous” at ${(x_{0},y_{0})}$ if

(i) ${f(x_{0},y_{0})}$ is defined and finite;

(ii) ${\displaystyle\lim_{(x,y)\rightarrow(x_{0},y_{0})}f(x,y)=f(x_{0},y_{0})}$ is defined;

(iii) ${\displaystyle\lim_{(x,y)\rightarrow(x_{0},y_{0})}f(x,y)}$ is defined (and finite).

Note: this can be determined by picking any curve ${\gamma\colon[0,1]\rightarrow{\mathbb R}^{2}}$ which satisfies
(4)$\displaystyle \gamma(t_{0}) = (x_{0},y_{0})$
for some ${0\leq t_{0}\leq 1}$, then taking
(5)$\displaystyle \lim_{t\rightarrow t_{0}}f\bigl(\gamma(t)\bigr) = \lim_{(x,y)\rightarrow(x_{0},y_{0})}f(x,y).$
The subtletly here lies with ${\gamma}$ being arbitrary. If two different curves produce two different results, the limit does not exist. Lets consider some examples and non-examples.

Example 1 (Limit Exists). Find
(6)$\displaystyle \lim_{(x,y)\rightarrow(2,4)}\frac{y+4}{x^{2}y-xy+4x^{2}-4x}.$

Solution: for this, we can simply plug in the values
(7)\displaystyle \begin{aligned} \lim_{(x,y)\rightarrow(2,4)}\frac{y+4}{x^{2}y-xy+4x^{2}-4x} &=\frac{(4)+4}{(2)^{2}(4)-(2)(4)+4(2^{2})-4(2)}\\ &=\frac{8}{16-8+16-8}=\frac{1}{2}. \end{aligned}
This is because the function is sufficiently nice.

Example 2 (Limit Doesn’t Exist). What is
(8)$\displaystyle \lim_{(x,y)\rightarrow(0,0)}\frac{x^{4}}{x^{4}+y^{2}}=?$

Solution: Lets first approach it along the ${x}$-axis, i.e. first setting ${y=0}$. We find
(9)$\displaystyle \lim_{(x,y)\rightarrow(0,0)}\frac{x^{4}}{x^{4}+y^{2}}=\lim_{x\rightarrow0}\frac{x^{4}}{x^{4}}=1.$
Now lets approach it on the ${y}$-axis, i.e. first setting ${x=0}$. We see
(10)$\displaystyle \lim_{(x,y)\rightarrow(0,0)}\frac{x^{4}}{x^{4}+y^{2}}=\lim_{y\rightarrow0}\frac{0}{0+y^{2}}=0.$
Still, approaching along the curve ${y=x^{2}}$ we see
(11)$\displaystyle \lim_{(x,y)\rightarrow(0,0)}\frac{x^{4}}{x^{4}+y^{2}}=\lim_{x\rightarrow0}\frac{x^{4}}{x^{4}+x^{4}}=\frac{1}{2}.$
But we have a problem: this implies ${0=1/2=1}$. This cannot be! So the limit cannot exist! Very sad.

3. Definition. Let ${z=f(x,y)}$ be defined on a region ${R}$ in the ${xy}$-plane, and let ${(x_{0},y_{0})}$ be an inerior point of ${R}$, we just don’t want a boundary point!

If
(12)$\displaystyle \lim_{\Delta x\rightarrow0}\frac{f(x_{0}+\Delta x,y_{0})-f(x_{0},y_{0})}{\Delta x}$
exists, then it is called the “Partial Derivative” of ${z=f(x,y)}$ at ${(x_{0},y_{0})}$ with respect to ${x}$. It is denoted
(13)$\displaystyle \frac{\partial}{\partial x}f = \frac{\partial}{\partial x}z =f_{x} = \partial_{x}f = \partial_{x}z$
evaluated at ${(x_{0},y_{0})}$. NB: the subscripts in the ${\partial_{x}}$ indicate what variable we are taking the partial derivative of, i.e., it’s shorthand for ${\partial_{x}=\partial/\partial x}$.

Under similar conditions,
(14)$\displaystyle \lim_{\Delta y\rightarrow0}\frac{f(x_{0},y_{0}+\Delta y)-f(x_{0},y_{0})}{\Delta y}$
is the partial derivative of ${z=f(x,y)}$ with respect to ${y}$ at ${(x_{0},y_{0})}$. We denote this by
(15)$\displaystyle \frac{\partial f}{\partial y}=\frac{\partial z}{\partial y}=\partial_{y}f = \partial_{y}z$
among a myriad of different conventions.

4. Higher order partial derivatives are done by taking it one at a time. So if
(16)$\displaystyle z=\mathrm{e}^{xy}$
for example, we have
(17)$\displaystyle \partial_{y}z=x\mathrm{e}^{xy}$
and taking its derivative again yields
(18)\displaystyle \begin{aligned} \partial_{y}^{2}z &= \partial_{y}\left(x\mathrm{e}^{xy}\right)\\ &=x\partial_{y}(\mathrm{e}^{xy}) \end{aligned}
Note we factor ${x}$ out in front of the partial derivative with respect to ${y}$ because ${x}$ is constant with respect to ${y}$. So we then obtain
(19)$\displaystyle \partial_{y}^{2}z = x^{2}\mathrm{e}^{xy}.$
We take partial derivatives one at a time, from right to left:
(20)$\displaystyle \partial_{x}\partial_{y}z = \partial_{x}\bigl(\partial_{y}z\bigr).$
Question: do partial derivatives commute? I.e., is ${\partial_{x}\partial_{y}=\partial_{y}\partial_{x}}$ always? Lets first consider an example calculation before considering an answer.

Example 3. Consider the function ${u=x^{2}-y^{2}}$. Find ${\partial_{x}^{2}u+\partial_{y}^{2}u}$.

Solution: We find that
(21)$\displaystyle \partial_{x}u = 2x\implies \partial_{x}^{2}u = 2.$
Similarly, we find
(22)$\displaystyle \partial_{y}u=-2y\implies \partial_{y}^{2}y=-2.$
Thus we conclude
(23)$\displaystyle \partial_{x}^{2}u+\partial_{y}^{2}u=2-2=0.$

5. Do Partial Derivatives Commute? Answer: not always. The conditions are fairly weak: if ${\partial_{x}z}$, ${\partial_{y}z}$, ${\partial_{x}\partial_{y}z}$ and ${\partial_{y}\partial_{x}z}$ are continuous throughout their respective domains, then
(24)$\displaystyle \partial_{x}\partial_{y}z = \partial_{y}\partial_{x}z.$

Exercise 1. Let ${u(x,t) = f(x+vt) + g(x-vt)}$ where ${v\not=0}$ is some constant. Prove
(25)$\displaystyle \partial_{t}^{2}u(x,t)=v^{2}\partial_{x}^{2}u(x,t).$

Exercise 2. Let ${f(x,y)=\ln(x^{2}+y^{2})}$. What is ${\partial_{x}^{2}f(x,y)+\partial_{y}^{2}f(x,y)}$?

Exercise 3. Consider ${g(x,y)=1/\sqrt{x^{2}+y^{2}}}$. What is ${\partial_{x}^{2}g(x,y)}$? What is ${\partial_{y}^{2}g(x,y)}$? Is ${\partial_{x}\partial_{y}g(x,y)=\partial_{y}\partial_{x}g(x,y)}$?

Exercise 4. Let ${f(x,y)=3x^{2}+4y^{3}++x^{2}y^{3}+\sin(xy)}$. What is ${\partial_{x}f}$? What is ${\partial_{y}f}$?

Exercise 5. Let ${z=\arctan(x^{2}\mathrm{e}^{2y})}$. What is ${\partial_{x}z}$? What is ${\partial_{y}z}$?