### Sufficient conditions for differentiability in multi-variable calculus.

We will be focusing on sufficient conditions of differentiability of $f:\Bbb{R}^2\to R$. The theorem says that if $f_x$ and $f_y$ exist and are continuous at point $(a,b)$, then $f$ is differentiable at $(a,b)$.

We have $f(x,y)$, which we know is partially differentiable with respect to $x$ and $y$, but may not be differentiable in general.

Differentiability at point $a$ in the $\Bbb{R}^n\to \Bbb{R}$ setting is described as $\frac{f(a+h)-f(a)}{\|h\|}-D \frac{h}{\|h\|}=\epsilon(h)$

$D$ is the matrix of partial derivatives with respect to the $n$ independent variables in $\Bbb{R}^n$.

What does all this mean? This is something that confused me for some time, and is likely to be helpful for others with the same doubts.

This is the $\epsilon-\delta$ definition of differentiation. We say $\lim\limits_{h\to 0} \frac{f(a+h)-f(a)}{h}$, if it exists, is the derivative of $f$ at $a$. Let the limit be $l$. We’re effectively saying thatÂ  for any $\epsilon>0$, $\left|\frac{f(a+h)-f(a)}{h}-l\right|<\epsilon$ if $0. Here $h=|x-a|$. However, this is not what we SEEM to say through this definition. What we seem to be saying is for any $\delta>0$ and $0, there exists $\epsilon>0$ such that $\frac{f(a+h)-f(a)}{\|h\|}-D \frac{h}{\|h\|}=\epsilon(h)$. Is there a difference? Yes. This will be illustrated below.

In our example of multi-variable differentiation, we’ve made $\epsilon$ a function of $h$. Why is that? What we mean by that is not that $\epsilon$ can be any function of $h$, like $h+c$, where $c$ is a constant. What we mean is $\lim\limits_{h\to 0} \epsilon=0$, although this is not obvious from the fact that $\epsilon$ is a function of $h$. Why should we explicitly mention the fact that $\epsilon$ should tend to $0$ as $h\to 0$? Because in the original definition $\frac{f(a+h)-f(a)}{\|h\|}-D \frac{h}{\|h\|}=\epsilon(h)$ we have made the argument that for any $\delta>0$ and $0 such that the difference between $\frac{f(a+h)-f(a)}{\|h\|}$ and $D \frac{h}{\|h\|}$ is $\epsilon$. Here, $\epsilon$ may not converge to $0$! We have just proven its existence, and none of its properties! For example, we could have said that for any $\delta>0$ and $0, $\frac{f(a+h)-f(a)}{\|h\|}-D \frac{h}{\|h\|}=2$. Here $\epsilon=2$. We have proven the existence of $\epsilon$.

It is only when we specify that $\lim\limits_{h\to 0} \epsilon=0$ that we make the definition of the derivative clear- that it is $\lim\limits_{h\to 0}\frac{f(a+h)-f(a)}{\|h\|}$. Such a limit is defined when for any $\epsilon>0$, there exists $\delta>0$ such that $\left |\frac{f(a+h)-f(a)}{\|h\|}-l\right|<\epsilon$ when $0.

Now we come back to sufficient conditions of differentiability. Let a function $f(x,y)$ be differentiable with respect to $x$ and $y$ at $(a,b)$. This implies

$f(a+\Delta x,b)-f(a,b)=\epsilon(\Delta x)\Delta x + f_x(a,b)\Delta x$

and

$f(a,b+\Delta y)-f(a,b)=\epsilon(\Delta y)\Delta y + f_y(a,b)\Delta y$.

Adding these two, we get

$f(a+\Delta x,b)+f(a,b+\Delta y)-2f(a,b)=\epsilon(\Delta x)\Delta x+\epsilon(\Delta y)\Delta y + f_x(a,b)\Delta x+f_y(a,b)\Delta y$.

Can we say

$f(a+\Delta x,b)-f(a,b)\approx f(a+\Delta x,b+\Delta y)-f(a,b+\Delta y)$,

assuming $\Delta x$ and $\Delta y$ are small enough? We have

$f(a+\Delta x,b)+f(a,b+\Delta y)-2f(a,b)=\epsilon(\Delta x)\Delta x+\epsilon(\Delta y)\Delta y+f_x(a,b)\Delta x+f_y(a,b)\Delta y$.

Now we use the property that the partial derivatives are continuous.

$\frac{f(a+\Delta x,b)-f(a,b)}{h}=\frac{f(a+\Delta x,b+\Delta y)-f(a,b+\Delta y)}{h}\implies f(a+\Delta x,b)-f(a,b)=f(a+\Delta x,b+\Delta y)-f(a,b+\Delta y)+o(h)$.

$o(h)$ is the correction factor which is utimately rectified. Remember that here $h=\Delta x$ or $\Delta y$. It can’t be a combination of both, as only partial derivatives $f_x,f_y$ are continuous.

We now have the formula

$f(a+\Delta x,b+\Delta y)=f_x(a,b)\Delta x+f_y(a,b)\Delta y+\epsilon(\Delta x)\Delta x+\epsilon(\Delta y)\Delta y+o(h)$.

The rest of the proof is elementary, and can be found in any complex anaysis textbook (pg. 67 of Complex Variabes and Applications, Brown and Churchill). I have only explained the difficult step in the proof.

Reiterating the theorem, if $f$ is is partially differentiable with respect to its independent variables at a particular point, and all those partial derivatives are continuous, then $f$ is differentiable at that point.