So...I am having trouble checking whether $F(x,y)=e^{x+y}$ is differentiable or not on $(0,0)$

The partial derivatives for $x$ and $y$ on $(0,0)$ (if no mistakes were made) are $1$ for both of them.

When checking the differentiability condition, I have:

$T_a(v)=1*\alpha + 1*\beta $


$\lim_{\alpha,\beta\rightarrow 0}\frac{f((0,0)+(\alpha,\beta))-f(0,0)-\alpha-\beta}{\sqrt{\alpha^2 +\beta^2}}$= $\lim_{\alpha,\beta\rightarrow 0}\frac{e^{\alpha+\beta}-1-\alpha-\beta}{\sqrt{\alpha^2 +\beta^2}}$= $\lim_{\alpha\rightarrow 0,\beta=m\alpha}\frac{e^{\alpha+m\alpha}-1-\alpha-m\alpha}{\sqrt{\alpha^2 +(m\alpha)^2}}$

Which ,in my case at least,will yield a result of the limit depending on $m$ and not being equal to $0$, thus showing that the function is not differentiable on $(0,0)$

This is part of an exercise which actually implies that, in fact, the function is differentiable on that point, so I am not sure which party made the mistake

Any help is appreciated.

  • $\begingroup$ Do you know that if a function has continuous partial derivatives at a point then it is differentiable at this point? $\endgroup$ – Mark Apr 18 '19 at 16:34
  • $\begingroup$ Yes, though I am not really sure how to prove it on paper (do I just take the non-evaluated partial derivative on each step and show that it's continuous?), which is why I am trying to show it with the limits $\endgroup$ – Lightsong Apr 18 '19 at 16:41
  • $\begingroup$ Well, both partial derivatives are equal to $e^{x+y}$ at each point $(x,y)\in\mathbb{R^2}$. And it is known that the exponential function is continuous everywhere, including at the point $(0,0)$. $\endgroup$ – Mark Apr 18 '19 at 16:58
  • $\begingroup$ Thanks! So just a quick proof of continuity on the individual, non-evaluated, partial derivatives is enough to get rid of the limit proof? I guess that with more complex functions just doing the limits on both sides is the fastest way to go? $\endgroup$ – Lightsong Apr 18 '19 at 17:10
  • $\begingroup$ If all partial derivatives are continuous at a point then the function is differentiable at this point. But the converse is not always true. There are differentiable functions for which the partial derivatives are not continuous. In such cases you have no choice, you must work with limits. $\endgroup$ – Mark Apr 18 '19 at 17:21

Since the partial derivatives of $f$ are continuous at $(0,0)$, $f$ is differentiable there (and, yes, $f'(0,0)(\alpha,\beta)=\alpha+\beta$). And, for each real $m$,$$\lim_{\alpha\to0}\frac{e^{\alpha+m\alpha}-1-\alpha-m\alpha}{\sqrt{\alpha^2+(m\alpha)^2}}=0.$$


Here's a way that avoids the direct calculation of partial derivaties: one "expects" the derivative to be $Df(\vec 0)(h,k)=h+k$, because $\nabla f(0)=(1,1).$

Now, if we check in the definition:

$f(h,k)-f(\vec 0)-(h+k)=e^{h+k}-1-(h+k)=$

$ 1+h+k+o(|h+k|^2)-1-(h+k)=o(|h+k|^2),$

we find that

$f(h,k)-f(\vec 0)-(h+k)=r(h,k)$ where $r=o(|h+k|^2)$

so our suspicion was correct.


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