Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

Thanks for your help.

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

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

share|improve this answer
    
(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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.