Obscurity Calculating Partial Derivative at a Point Using Limit Definition

Question

In the beginning of a chapter that speaks about second order directional derivatives, my textbook presents the following example that shows that generally, the order by which we derive a certain function may matter greatly. Unfortunately I got stumped by some step along the way. The example goes as follows:

$$ f:\mathbb{R}^2 \rightarrow \mathbb{R} $$ $$ (x,y) \longrightarrow \begin{cases} \frac{x^2 - y^2}{x^2 + y^2}xy & (x,y)\neq(0,0)\\ 0 & (x,y)=(0,0) \end{cases} $$ As the function is broken up at the point $(0,0)$ we must use the limit definition to calculate the partial derivatives at $(0,0)$. So, the way I thought it should be done is by calulating as follows: $$ \frac{f_{|(0,0),(1,0)}(t) - f_{|(0,0),(1,0)}(0)}{t}=\frac{\frac{t^2 - 0}{t^2}t\cdot0 - 0}{t} = 0\xrightarrow[t \rightarrow 0]{}0 $$ I thought that at this point we can safely say $\frac{\partial f}{\partial x}(0,0) = 0$. However, my textbook, though it starts as I did, goes further to say: $$ \text{when } y \neq 0 \text{ we have } \frac{f(x,y) - f(0,y)}{x - 0} = \frac{x^2 - y^2}{x^2 + y^2}y \xrightarrow[x \rightarrow 0]{}-y $$

And thus we have $\frac{\partial f}{\partial x}(0,y) = -y$. Next the second order is calculated: $\frac{\partial^2 f}{\partial y \partial x}(0,0) = -1$. The same process is repeated in the reverse order and we get $\frac{\partial^2 f}{\partial x \partial y}(0,0) = 1$.

I went over the definitions of directional derivative and partial derivative yet I still don't fully understand the process. Why do we calculate $\frac{\partial f}{\partial x}(0,y)$, that is for all $y$? Isn't $\frac{\partial f}{\partial x}(0,0)$ suppose to be enough to find the second order in the $y$ direction?

Thanks!

Answer 1

I think your confusion is coming from the wrong idea that $\frac{\partial^2 f}{\partial y \partial x}(0,0) = \frac{\partial}{\partial y}(\frac{\partial f}{\partial x}(0,0)).$

Rather, in order to find $\frac{\partial^2 f}{\partial y \partial x}(0,0)$, we proceed as follows. For every fixed real $y=\eta$, we show that the fixed number $\frac{\partial f}{\partial x}(0,\eta)$ exists. This is exactly your first two calculations. Note that $\frac{\partial f}{\partial x}(0,\eta)$ is $\frac{d}{dx}[f(x,\eta)]$ evaluated at $x=0$.

Actually all we need is the function of $y$ given by $\frac{\partial f}{\partial x}(0,y)$ to be defined in some neighborhood $(-\epsilon,\epsilon)$. It is, by the work you showed.

$\frac{\partial^2 f}{\partial y \partial x}(0,0)$ is $\frac{d}{dy}(\frac{\partial f}{\partial x}(0,y))$ evaluated at $y=0$.

That is, $\frac{\partial^2 f}{\partial y \partial x}(0,0)$ is $\frac{d}{dy}(-y)$ evaluated at $y=0$.

Answer 2

Your function $f$ is a classic example. You should contrast it with $$ G(x,y) = \begin{cases} xy &\,\,\,\, |y|\leq|x|\\ -xy &\,\,\,\, |y|>|x|. \end{cases} $$

The point of each example is to show that, in your words, "the order may matter greatly"; this despite the fact that each of $f$ and $G$ is continuous at the origin. One of several impressive things about $f$ is that both (first-order) partial derivatives are continuous at the origin, as functions of two variables. (You could investigate if this is also true for $G$.)

As for your question, I think you answered it yourself. The derivative of $-y$ with respect to $y$ is $-1$.

And thus we have $\frac{\partial f}{\partial x}(0,y) = -y$. Next the second order is calculated: $\frac{\partial^2 f}{\partial y \partial x}(0,0) = -1$.

The object $\frac{\partial f}{\partial x}(0,0)$ is just a number, a constant! You need a function of an independent quantity (for us, $y$) to find a derivative with respect to $y$.

$\frac{\partial f}{\partial x}(0,0)$ alone is not useful for finding a second-order partial.