# Multivariable differentiability: where does the concept come from?

I'm not sure if I posed the right question, but this is my curiosity:

That a function is differentiable in $P\in\mathbb{R}^n$ means that given $F:\mathbb{R}^n\rightarrow\mathbb{R}^m$

$$\lim_{X\to P}\frac{\|F(X)-F(P)-J(X-P)\|}{\|X-P\|}=0$$ I understand that in terms of functions that go from $\mathbb{R}^n\to\mathbb{R}$, that expression basically means that given a hyperplane that goes through the point $(P,f(P))$, as you approach that point along any curve of points $(X,f(X))$, the angle of the triangle with those two points as vertices and the coordinate of the plane in $X$ should approach $0$. That at least gives me an intuition of how the plane is actually approximating the function near that point. But,

$1)$ Why is that the definition of differentiation, exactly?

$2)$ Is there an intuition for the more general $\mathbb{R}^n\rightarrow\mathbb{R}^m$ type of function?

• "Is there an intuition for the more general Rn→Rm type of function?" Consider a ball (sphere with center P) in the domain with radius decreasing to zero. Evaluating f any path to P inside the ball should have a limit of f(P) – nickalh May 8 '15 at 10:08

## 1 Answer

Concerning functions, that is $f:\mathbb R^n\to\mathbb R$, I'd rather put it this way. One understands derivation in one variable, hence starts by derivating the restrictions $f|\ell$ to lines $\ell$ through a point $P$. This is just a one variable limit. Then, reasonably enough one asks these limits exist for all lines $\ell$ defined by a non-zero vectors $u$ at the point (for the limit definition it is not necessary the vector to be unitary, and if $u=0$ the limit is trivially $0$). This are the directional derivatives $D_uf(P)$, which give the map $J(u)=D_uf(P)$. Geometrically, each directional derivative gives the (tangent) line that best approximate the section of the graph $\varGamma\subset\mathbb R^n\times\mathbb R$ of $f$ cut by the plane $\ell\times\mathbb R$. But nothing guarantees all these tangent lines form a hyperplane! Thus why one asks the function $J:\mathbb R^n\to\mathbb R$ (with $L(0)=0$) to be linear. But still one knows that in several variables a function can have a limit along all directions without having a global limit, hence one asks for the global limit condition $$\lim_{u\to0}\frac{f(P+u)-f(P)-J(u)}{\|u\|}=0.$$ For a mapping $f:\mathbb R^n\to\mathbb R^m$ the variation is that the sections of the graph are by $(m+1)$-planes $\ell\times\mathbb R^m$ and you get more arbitrary curves (instead of planar). Still the derivative of that restriction gives the tangents to those curves. But this enters more the area of a curves and surfaces course. From the Analysis viewpoint I'd resource to the work componentwise general idea.

• I understand, that means every directional derivative must lie on the plane. I didn't really understand the second part. I know I could just deal with components, but then what is the reason for the differentiation to be defined that way in those functions? I mean, it's obvious it must be analogous to the other ones, but in which way? I don't know if I'm being clear enough. I obviously don't need to know this for the course I'm taking, but I'm curious. Anywhere to read on the subject? – sbs95 Feb 18 '15 at 10:36
• One way to think of it is that the jacobian matrix consists of the gradients of the components, and one single formula gathers all the information. Geometrically, it's better to look a the case $f:\mathbb R\to\mathbb R^m$. Such a function is a differential curve in $\mathbb R^m$, and then $f'(P)=Jf(P)\in\mathbb R^m$ is the velocity of the curve at $P$, or the tangent vector at $f(P)$. I think the geometric ideas behind differentiation come out best after the Implicit Functions Thm and regular subvarieties and Lagrange multipliers are studied. – Jesus RS Feb 18 '15 at 11:13