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Let $f:\mathbb{R}^2\to\mathbb{R}$ and $(p,q)\in\mathbb{R}^2$ such that both $f_x$ and $f_y$ exist at $(p,q)$. Suppose $f_x$ is continuous at $(p,q)$, show that the directional derivative $D_vf$ at $(p,q)$ exists for all (nonzero) $v\in\mathbb{R}^2$.

My attempt:

$f_x(p,q)=\lim_{h\to 0}\frac{f(p+h,q)-f(p,q)}{h}$ exists.

$f_y(p,q)=\lim_{h\to 0}\frac{f(p,q+h)-f(p,q)}{h}$ exists.

Let $v=(v_1,v_2)\in\mathbb{R}^2$.

$\begin{align*} D_vf(p,q)&=\lim_{h\to 0}\frac{f(p+hv_1,q+hv_2)-f(p,q)}{h}\\ &=\lim_{h\to 0}\frac{f(p+hv_1,q+hv_2)-f(p,q+hv_2)}{h}+\lim_{h\to 0}\frac{f(p,q+hv_2)-f(p,q)}{h} \end{align*}$

I am stuck here, and I am not entirely sure this is the right track to begin with.

Thanks for any help!

yoyostein
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  • I think this will help http://math.stackexchange.com/questions/372650/continuity-of-one-partial-derivative-implies-differentiability –  May 18 '16 at 06:21
  • For my question, the partial derivative is only continuous at a single point. – yoyostein May 18 '16 at 06:39

1 Answers1

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You are off to a good start. For the second term, just multiply top and bottom by $v_2$ and then use the existence of $f_y$. Treat $v_2=0$ separately. For the first term, use the mean value theorem to find a $c$ that lies between $p$ and $p+hv_1$ so you can write

$$ f(p+hv_1,q+hv_2) - f(p,q+hv_2) = f_x(c,q+hv_2)hv_1.$$

Then use the continuity of $f_x$ to evaluate the limit. Note that $c\to p$ as $h\to{} 0$.

Edit:

Because $f_x$ is continuous at the point, it must exist on an open interval around the point. (keeping $y $ fixed for a moment). That is, $f $ is differentiable on an open interval and hence is continuous on a smaller closed interval. This is what's required to apply the mean value theorem.

hilbert spaceman
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