### Multi-variable differentiation.

There are very many bad books on multivariable calculus. “A Second Course in Calculus” by Serge Lang is the rare good book in this area. Succinct, thorough, and rigorous. This is an attempt to re-create some of the more orgasmic portions of the book.

In $\Bbb{R}^n$ space, should differentiation be defined as $\lim\limits_{H\to 0}\frac{f(X+H)-f(X)}{H}$? No, as division by a vector $(H)$ is not defined. Then $\lim\limits_{\|H\|\to 0}\frac{f(X+H)-f(X)}{\|H\|}$? We’re not sure. Let us see how it goes.

Something that is easy to define is $f(X+H)-f(X)$, which can be written as $f(x_1+h_1,x_2+h_2,\dots,x_n+h_n)-f(x_1,x_2,\dots,x_n)$ ( $H$ is the $n$-tuple $(h_1,h_2,\dots,h_n)$).

This expression in turn can be written as $f(x_1+h_1,x_2+h_2,\dots,x_n+h_n)-f(x_1,x_2,\dots,x_n)=\left[f(x_1+h_1,x_2+h_2,\dots,x_n+h_n)-f(x_1,x_2+h_2,\dots,x_n+h_n)\right]\\+\left[f(x_1,x_2+h_2,\dots,x_n+h_n)-f(x_1,x_2,\dots,x_n+h_n)\right]+\dots+\left[f(x_1,x_2+h_2,\dots,x_n+h_n)-f(x_1,x_2+h_2,\dots,x_n)\right]$.

Here, we can use the Mean Value Theorem. Let us suppose $s_1\in((x_1+h_1,x_2+h_2,\dots,x_n+h_n),(x_1,x_2+h_2,\dots,x_n+h_n))$,

or in general $s_k\in((x_1,x_2,\dots,x_k+h_k,\dots,x_n+h_n),(x_1,x_2,\dots,x_k\dots,x_n+h_n))$. Then $f(x_1+h_1,x_2+h_2,\dots,x_n+h_n)-f(x_1,x_2,\dots,x_n)=\\ \displaystyle{\sum\limits_{k=1}^n{D_{x_k}(x_1,x_2,\dots,s_k,\dots,x_n+h_n).((x_1,x_2,\dots,x_k+h_k,\dots,x_n+h_n)-(x_1,x_2,\dots,x_k,\dots,x_n+h_n))}}$.

No correction factor. Just this.

What follows is that a function $g_k=D_{x_k}(x_1,x_2,\dots,s_k,\dots,x_n+h_n)-D_{x_k}(x_1,x_2,\dots,x_k,\dots,x_n)$

is assigned for every $k=\{1,2,3,\dots,n\}$.

Hence, the expression becomes $f(x_1+h_1,x_2+h_2,\dots,x_n+h_n)-f(x_1,x_2,\dots,x_n)=\sum\limits_{k=1}^n {D_{x_k}(x_1,x_2,\dots,x_n)+g_k}$

It is easy to determine that $\lim\limits_{H\to 0}g_k=0$.

The more interesting question to ask here is that why did we use mean value theorem? Why could we not have used the formula $f(x_1+h_1,x_2+h_2,\dots,x_n+h_n)-f(x_1,x_2,\dots,x_n)\\=\sum\limits_{k=1}^n {\left[D_{x_k}(x_1,x_2,\dots,x_k\dots,x_n+h_n)+g_k(x_1,x_2,\dots,x_k,\dots,x_n+h_n,h_k)\right]}$,

where $\lim\limits_{h_k\to 0}g_k(x_1,x_2,\dots,x_k,\dots,x_n+h_n,h_k)=0$??

This is because $g_k(x_1,x_2,\dots,x_k,\dots,x_n+h_n,h_k)$ may not be defined at the point $(x_1,x_2,\dots,x_n)$. If in fact every $g_k$ is continuous at $x_1,x_2,\dots,x_n)$, then we wouldn’t have to use mean value theorem.

Watch this space for some more expositions on this topic.

A function is differentiable at $X$ if it can be expressed in this manner: $f(X+H)-f(X)=($grad $f(X)).H+\|H\|g(X,H)$ such that $\lim\limits_{\|H\|\to 0}g(X,H)=0$. This is a necessary and sufficient condition; the definition of differentiability. It does not have a derivation. I spent a very long time trying to derive it before realising what a fool I had been.