# Differentiability VS derivability

1. Feb 4, 2006

### quasar987

differentiability VS "derivability"

In french, the quality of a function which in english is called 'differentiable', we call 'dérivable'. And we call 'différentiable' at the point (x,y) a function f such that we can write f(x+h,y+k) - f(x,y) = h*df(x,y)/dx + k*df(x,y)/dy + o(sqrt{h²+k²}).

What is this attribute called in english?

2. Feb 5, 2006

### Tide

Such a function is called "analytic."

3. Feb 5, 2006

### Hurkyl

Staff Emeritus
No, that's just the criterion for the function to be differentiable at the point (x,y). Analytic is much stronger.

4. Feb 5, 2006

### quasar987

My teacher insisted that the notion of "différentiabilité" was stronger than that of "dérivabilité".

5. Feb 5, 2006

### Hurkyl

Staff Emeritus
Being differentiable at (x,y) is stronger than having both of its partial derivatives existing at (x,y). Maybe this is to what your teacher is referring?

In English, a function is differentiable at x (=(x, y)) iff there exists a linear transformation T such that:

$$\lim_{\vec{v} \rightarrow 0} \frac{||f(\vec{x} + \vec{v}) - f(\vec{x}) - T\vec{v}||}{||\vec{v}||} = 0$$

In the case at hand, T is just the gradient of f at x, so this works out to:

$$\lim_{(h,k) \rightarrow 0} \frac{|f(x+h, y+k) - f(x, y) - h \frac{\partial f}{\partial x}(x,y) - k \frac{\partial f}{\partial y}(x, y)|}{\sqrt{h^2 + k^2}} = 0$$

which is equivalent to your expression.

Last edited: Feb 5, 2006
6. Feb 5, 2006

### quasar987

Well, my mistake. I was certain that in english, differentiable was a synomym of "both of its partial derivatives exist".

7. Feb 5, 2006

### Hurkyl

Staff Emeritus
Ugh, it's been a while since I've considered these edge conditions, so I might have this wrong:

I think "f is differentiable" is logically equivalent to "f's partial derivatives exist, and are continuous", so you we're almost right.

(If it makes you feel better, it took me a little while before I could remember for sure that I had the definitions right!)

8. Feb 5, 2006

### mathwonk

if both partials exist and are continuous, then the function is (frechet) differentiable, but not vice versa.

i seem to recall that in some old books the term "differentiable" was used to refer to existence of partials as opposed to frechet derivatives.

remember that every author sets his own terminology.

if one uses continuous derivtives then the existence of partials is equivalent to frechet derivatives,

i.e. to say the frechet derivatives exist and define continuois mapsinto the space of linear functions, is equivalent to saying the partials exist and are continuous.

as always, a very useful source for basic calculus including all these facts is the outstanding book of dieudonne, foundations of modern analysis.

Last edited: Feb 5, 2006