# Differentiability and continously differentiable definition/concepts.

1. May 16, 2014

### binbagsss

Theorem: ctsly differentiable at a if the function is cts and its partial derivatives exist and are cts in a neighborhood of a. [1]

- so to be differentiable we can check whether this conditions holds, and if it does ctsly diff => diff.

- the definition of a scalar function being differentiable at the point a is f(a+h)-f(a)=h.v(a)+R(h)... [2] ; for a scalar function of f.

and lim $_{h \rightarrow 0}$ $\frac{R(h)}{ h }$ = 0 [3]

(sorry this should be modulus h . I cant get it to work ! )

- BUT, if this doesn't hold, then we can go back to the scalar differentiable definition and check if R(h) obeys condition [3]

Questions:

- when we deduce what R(h) is, what should we take ∇f as - should it be the value you get from the partial derivatives (limit definition), or from 'directly differentiating' f. (so this would assume that the partial derivative does exist , and failed on theorem [1] condition by the partials not being cts, being the reason I am looking back at definition [2]).

- The definition of cts, is, that the limit needs to exist in a neighborhood of point a, and not at the point a. So if condition [1] fails on the partial derivative not existing at a , this should not matter? we just need to check they are cts in a neighborhood of a? And ctsly differentiable is still a possibility? ( I ask because my solutions always seem to take the partial at the point a, or would this be more for ∇f?

- cts diff => diff. condition [1]. This does not work the other way around , so differentiability is still a possibility. Am I correct in thinking that a function can still be differentiable if:
a) its partial derivative does not exist at a
b) they are not cts in a neighborhood of a

- but regarding a) , if the partial derivatives do not exist, from [2] the only candidate for v(a) is ∇f , which is attained from the partial derivatives , so if these do not exist, as a limit, (the partial derivatives can not be cts) and we must get the partial derivatives from the function without the limit definition ?

(in the case that they are cts , this limit should equal the partial derivatives attained from the function evaluated at this point, so you take either for ∇f (as they are same ) - is this correct? (I Know you wouldn't need to go back to the definition in this case as theorem [1] conditions are met, but I'm checking my understanding..

Many Thanks for any assistance, greatly appreciated !

2. May 16, 2014

### SammyS

Staff Emeritus
What are these abbreviations ?

3. May 17, 2014

### binbagsss

Which ones? Ctsly = continuously, and cts= continuous. Any others?

4. May 17, 2014

### SammyS

Staff Emeritus
Those are the two in particular which encouraged me to not read through the Original Post .

I suspect that other helpers on the forums may have the same reaction. Those who generally chime in on such subjects haven't responded to this thread so far.

5. May 17, 2014

### micromass

Staff Emeritus
The idea is to take the partial derivatives and use them to find the gradient of $f$. That is, we have

$$\nabla f(a) = (\frac{\partial f}{\partial x_1} (a), ..., \frac{\partial f}{\partial x_n}(a))$$

Use this to define $\nabla f$. Then you can check differentiability using this definition.
You know that if $f$ is differentiable, then $\nabla f$ has the previous form, so it can't be anything else.

You need to find the partial derivatives at every point possible, not only at $a$. You will need the partial derivatives to exist at some neighborhood of $a$ and you will need them to be continuous there. Just checking continuity at one single point is not sufficient.

No. If a function is differentiable, then its partial derivatives must exist.

That can happen. It can happen that the function is differentiable but that the partials don't exist.

6. May 19, 2014

### binbagsss

What if, for example, you have f = lx^2-y^2 l and are investigating differentiability at the origin. for
x^2-y^2 > 0 we get f= x^2-y^2 ,
for
x^2-y^2<0 we get f= -x^2+y^2.

So how do we 'read of/apply the gradient function' to the origin ? If we did the limit partial deritivate definition, then we can keep the mod signs , and don't need to decide? In this case isn't the only way to get ∇f, by partial deriviative definition?

7. May 19, 2014

### micromass

Staff Emeritus
First you'll need to find the partial derivatives at the origin. Then you can use that to make a candidate for the gradient $\nabla f$.

8. May 19, 2014

### binbagsss

so you use the limit definition?

9. May 19, 2014

### micromass

Staff Emeritus
In this case, yes.