Check whether $F(x,y)$ is differentiable or not at $(0,0)$

Question

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 $

$||v||=\sqrt{\alpha^2+\beta^2}$

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

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.

Answer 1

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 $m$,$$\lim_{\alpha\to0}\frac{e^{\alpha+m\alpha}-1-\alpha-m\alpha}{\sqrt{\alpha^2+(m\alpha)^2}}=0.$$

Answer 2

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.