# Question on differentiability at a point

I have a question that asks me to show that a function $f(x,y)$ is differentiable at the point $(x_0,y_0)$ if given that $\partial f\over\partial x$ is continuous in an open ball around $(x_0,y_0)$ and $\partial f\over\partial y$ exists at that point. Based on the given information, I assume the continuity of $\partial f\over\partial x$ in the open ball is necessary, but doesn't the existence of $\partial f\over\partial x$ and $\partial f\over\partial y$ at that point mean that $f$ is differentiable there? How does that "extra" information come into play? ::confused::

-
Existence of the partials at a point does not imply differentiablity. –  David Mitra Nov 22 '11 at 10:28
@DavidMitra: Thanks, I have inferred that from the question, but I don't understand why that is so. –  ringo Nov 22 '11 at 10:54
Take $f(x,y)=\cases{ {xy\over x^2+y^2}, &(x,y)\ne(0,0)\cr 0,&(x,y)=(0,0)}$. $f_x$ and $f_y$ exist at $(0,0)$, but $f$ is not even continuous at $(0,0)$. (Look at what $f$ does on the $x$ and $y$ axes and on the line $x=y$.) –  David Mitra Nov 22 '11 at 11:02

Remember the definition of differentiability:

The function $f(x,y)$ is differentiable at $(x_0,y_0)$ if there exists a linear functional $L$ such that for every $(x_1,y_1)$ $$\lim_{t\to 0} \frac{f(x_0+ tx_1, y_0 + t y_1) - f(x_0,y_0) - tL(x_1,y_1)}{t} = 0$$

The existence of $f_x$ only guarantees that the above is true for all $(x_1,0)$, and the existence of $f_y$ only guarantees that it is true for all $(0,y_1)$, and they are far from the whole set of $(x_1,y_1)$.

For this problem, you need to do the following. First let $L(a,b) = a f_x(x_0,y_0) + b f_y(x_0,y_0)$. Next observe that

$$f(x_0 + t x_1, y_0 + t y_1) - f(x_0,y_0) = f(x_0 + t x_1,y_0 + t y_1) - f(x_0, y_0 + t y_1) + f(x_0, y_0 + t y_1) - f(x_0,y_0)$$

The first two terms on the right hand side differs from $tx_1f_x(x_0, y_0 + ty_1)$ by a term of $O(t^2)$. And $tx_1f_x(x_0,y_0+ty_1)$, using continuity of $f_x$, differs from $tL(x_1,0)$ by a term that is $o(t)$. The last two terms differ from $tL(0,y_1)$ by a term of $O(t^2)$ using the definition of partial derivatives.

Putting this all together you will get differentiability.

What is the key in the above step? The idea is the following: to prove differentiability, you need to show that you can take an infinitesimal step in any direction, starting from $(x_0,y_0)$. Having $f_x$ and $f_y$ at the point $(x_0,y_0)$ shows that you can go "horizontally" (in the $x$ direction) or "vertically" (in the $y$ direction). But just having those two directions you cannot go in a "diagonal" direction, so that does not suffice to get differentiability. (Imagine standing at the mountain top. You see that in front of you there is a path along the ridge, and to you left there is a path along the ridge. You shouldn't start walking in the 11 o'clock direction because there may actually be a cliff there.)

But if you have that $f_x$ is continuous on an entire neighborhood of $(x_0,y_0)$, you can go "diagonally" by first going vertically (using that $f_y$ exists at $(x_0,y_0)$), then going "horizontally" (using that $f_x$ exists at $(x_0, y_0 + t y_1)$).

-
Thank you, Willie! Great analogy! –  ringo Nov 22 '11 at 17:11