We studied differentials in Section 4.4, where Definition 4.4.2 states that if $y=f(x)$ and $f$ is differentiable, then $dy=\fp(x)dx\text{.}$ One important use of this differential is in Integration by Substitution. Another important application is approximation. Let $\dx = dx$ represent a change in $x\text{.}$ When $dx$ is small, $dy\approx \dy\text{,}$ the change in $y$ resulting from the change in $x\text{.}$ Fundamental in this understanding is this: as $dx$ gets small, the difference between $\dy$ and $dy$ goes to 0. Another way of stating this: as $dx$ goes to 0, the error in approximating $\dy$ with $dy$ goes to 0.

We extend this idea to functions of two variables. Let $z=f(x,y)\text{,}$ and let $\dx = dx$ and $\dy=dy$ represent changes in $x$ and $y\text{,}$ respectively. Let $\ddz = f(x+dx,y+dy) - f(x,y)$ be the change in $z$ over the change in $x$ and $y\text{.}$ Recalling that $f_x$ and $f_y$ give the instantaneous rates of $z$-change in the $x$- and $y$-directions, respectively, we can approximate $\ddz$ with $dz = f_xdx+f_ydy\text{;}$ in words, the total change in $z$ is approximately the change caused by changing $x$ plus the change caused by changing $y\text{.}$ In a moment we give an indication of whether or not this approximation is any good. First we give a name to $dz\text{.}$

We can approximate $\ddz$ with $dz\text{,}$ but as with all approximations, there is error involved. A good approximation is one in which the error is small. At a given point $(x_0,y_0)\text{,}$ let $E_x$ and $E_y$ be functions of $dx$ and $dy$ such that $E_xdx+E_ydy$ describes this error. Then $$\begin{align*} \ddz \amp = dz + E_xdx+ E_ydy\\ \amp = f_x(x_0,y_0)dx+f_y(x_0,y_0)dy + E_xdx+E_ydy. \end{align*}$$

If the approximation of $\ddz$ by $dz$ is good, then as $dx$ and $dy$ get small, so does $E_xdx+E_ydy\text{.}$ The approximation of $\ddz$ by $dz$ is even better if, as $dx$ and $dy$ go to 0, so do $E_x$ and $E_y\text{.}$ This leads us to our definition of differentiability.

Our intuitive understanding of differentiability of functions $y=f(x)$ of one variable was that the graph of $f$ was “smooth.” A similar intuitive understanding of functions $z=f(x,y)$ of two variables is that the surface defined by $f$ is also “smooth,” not containing cusps, edges, breaks, etc. The following theorem states that differentiable functions are continuous, followed by another theorem that provides a more tangible way of determining whether a great number of functions are differentiable or not.

The theorems assure us that essentially all functions that we see in the course of our studies here are differentiable (and hence continuous) on their natural domains. There is a difference between Definition 12.4.3 and Theorem 12.4.6, though: it is possible for a function $f$ to be differentiable yet $f_x$ and/or $f_y$ is not continuous. Such strange behavior of functions is a source of delight for many mathematicians.

When $f_x$ and $f_y$ exist at a point but are not continuous at that point, we need to use other methods to determine whether or not $f$ is differentiable at that point.

For instance, consider the function $$\begin{equation*} f(x,y) = \left\{\begin{array}{cl} \frac{xy}{x^2+y^2} \amp (x,y)\neq (0,0) \\ 0 \amp (x,y) = (0,0) \end{array} \right. \end{equation*}$$

We can find $f_x(0,0)$ and $f_y(0,0)$ using Definition 12.3.2: $$\begin{align*} f_x(0,0) \amp = \lim_{h\to 0} \frac{f(0+h,0) - f(0,0)}{h}\\ \amp = \lim_{h\to 0} \frac{0}{h^2} = 0;\\ f_y(0,0) \amp = \lim_{h\to 0} \frac{f(0,0+h) - f(0,0)}{h}\\ \amp = \lim_{h\to 0} \frac{0}{h^2} = 0. \end{align*}$$

Both $f_x$ and $f_y$ exist at $(0,0)\text{,}$ but they are not continuous at $(0,0)\text{,}$ as $$\begin{equation*} f_x(x,y) = \frac{y(y^2-x^2)}{(x^2+y^2)^2} \qquad \text{ and } \qquad f_y(x,y) = \frac{x(x^2-y^2)}{(x^2+y^2)^2} \end{equation*}$$ are not continuous at $(0,0)\text{.}$ (Take the limit of $f_x$ as $(x,y)\to(0,0)$ along the $x$- and $y$-axes; they give different results.) So even though $f_x$ and $f_y$ exist at every point in the $x$-$y$ plane, they are not continuous. Therefore it is possible, by Theorem 12.4.6, for $f$ to not be differentiable.

Indeed, it is not. One can show that $f$ is not continuous at $(0,0)$ (see Example 12.2.10), and by Theorem 12.4.5, this means $f$ is not differentiable at $(0,0)\text{.}$