# Partial derivatives and differentiable function

I'm trying to show:

Let $f:A\subset \mathbb{R}^2 \to \mathbb{R}$, if the partial derivative $d_{e1}f$ exists and is continuous in open set $V\subset A$ around $x_0$ and $d_{e2}f$ exists, then $f$ is differentiable in $x_0$.

I know that if the partial derivatives exists and it's continuous then $f$ is differentiable, but I can't use this result.

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By the mean value theorem, \begin{align*} f(x_0 + h, y_0 + k) - f(x_0, y_0) &= f(x_0 + h, y_0 + k) - f(x_0, y_0 + k) + f(x_0, y_0 + k) - f(x_0, y_0)\\ & = hD_x f(x_0 + \xi, y_0 + k) + f(x_0, y_0 + k) - f(x_0, y_0) \end{align*} (For some $\xi$ between $0$ and $h$.)

Hence:

\begin{align*} &\frac{f(x_0 + h, y_0 + k) - f(x_0, y_0) - hD_x f(x_0, y_0) - k D_y f(x_0, y_0)}{\sqrt{h^2 + k^2}}\\ &= \frac{h}{\sqrt{h^2 + k^2}}(D_x f(x_0 + \xi, y_0 + k) - D_x f(x_0, y_0))\\ &\quad + \frac{k}{\sqrt{k^2 + h^2}}\left(\frac{f(x_0, y_0 + k) - f(x_0, y_0)}{k} - D_y f(x_0, y_0) \right) \end{align*}

If you take the limit as $(h, k) \to 0$, the remaining fractions are bounded, the first bracket vanishes by continuity, and the second by definition. QED

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(I'm sorry I don't know how to make line breaks in math mode...) – Tom Bachmann Sep 20 '11 at 20:15
I fixed it for you. – Arturo Magidin Sep 20 '11 at 20:26
Ah I see. I was not aware of the extent of latex compatiblity in matjax. cool; thanks. – Tom Bachmann Sep 20 '11 at 20:56
@Tom Bachmann: Thanks! – Hiperion Sep 20 '11 at 21:45