# Problem on differentiability

1. Oct 14, 2013

### mahler1

The problem statement, all variables and given/known data.

Let $f:\mathbb ℝ:→ℝ^2$ be a function defined as:
$f(x,y)=\frac{x^2y-2xy+y} {(x-1)^2+y^2} \forall (x,y)≠(1,0)$ and $f(1,0)=0$.
Prove that for any curve $α:(-ε,ε)→ℝ^2$ of class $C^1$ (where $ε>0$) such that $α (0)=(1,0)$ and $α(t)≠(1,0)$ for every $t≠0$, the derivative $(foα)'(0)$ exists; but that $f$ is not differentiable at the point $(1,0)$.

The attempt at a solution:

I didn't have any problems to prove that f isn't differenitable at $(1,0)$. First, I proved that the partial derivatives exist. Both of them gave 0, then supposed $f$ was differentiable at that point. If this was the case, the limit $\lim_{(x,y)\rightarrow (1,0)} \frac{ |f(x,y)-(f(1,0)+<∇f(1,0),(x-1,y)>|} {||(x-1,y)||}$ would have to equal 0. If I consider the curve $ψ(t)=(t+1,t)$, then the limit is $\frac{\sqrt(2)} {4}≠0$.

I got stuck in proving the first part of the statement "for any curve $α:(-ε,ε)→ℝ^2$..., the derivative $(foα)'(0)$ exists". I mean, I can't show this using the limit definition: $\lim_{h\rightarrow 0} \frac {foα(h)-foα(0)} {h}$, because I don't have enough information to calculate it. Am I doing something wrong? What should I do to prove that part of the statement?

Last edited: Oct 14, 2013
2. Oct 14, 2013

### brmath

You know what f(x,y) is and you have a(t) = (x(t), y(t)). So you are looking at f(x(t),y(t)). Differentiate with respect to t, being careful about the chain rule. You will get some expression involving derivatives coming from f which you can see are okay, and a'(t) which by definition was $C^1$.

3. Oct 14, 2013

### mahler1

${\frac{\partial f}{\partial y}}(1,0)$ and ${\frac{\partial f}{\partial x}}(1,0)$ exist and are both $0$. So, what you've suggested is to apply the chain rule. By the chain rule $(foα)'(0)=<∇f(α(0)),α'(0)>=<(0,0),(x'(t),y'(t)>=0$ Is this ok? I thought that in order to apply the chain rule I needed both $f(x,y)$ and $α(t)$ to be differentiable, that's why I didn't try to solve it this way earlier. Instead, I've tried to proved differentiability by using the limit definition.

4. Oct 14, 2013

### Dick

f isn't differentiable in the two dimensional sense of the word at (1,0). But it does have well defined directional derivatives. Another way to think about this is that if you take a C^1 curve then the derivative along the parameter t is the same as the derivative along a tangent line to the curve at (1,0). Just check all tangent lines through (1,0) have a well defined derivative at (1,0).

5. Oct 15, 2013

### brmath

Re the chain rule, it applies fine to partial derivatives and is used that way all the time.

Your computation looks plausible but I think it is not quite correct. You have shown that the expression $∇f(a(0)) \cdot a'(0)$ exists and is 0. This is not the same as showing the $\lim \frac {∇f(a(t)) \cdot a'(t) - ∇f(a(0)) \cdot a'(0)}{t}$ exists. There might be some function f where $∇f(a(0)) \cdot a'(0)$ = 0, and yet the limit itself might not exist.

Unless you are prepared to show that scenario cannot happen -- and I suspect that it can -- you need to look specifically at the f you were given, and work out the f'(a(t)) based on that function. Then take the limit as t $\rightarrow$ 0.

Your expression $\psi(t)$ is a perfect example of the kind of a(t) you are looking at. In this case you can see that f($\psi(t)$) has a derivative as a function of t at 0. The purpose of the a(t) is to analyze the situation in general.

I think what this problem is getting at is that having partial derivatives and being differentiable are not the same thing in any dimension > 1, and you seem to have grasped that at least in terms of formally writing down your definition of differentiable.

What differentiable at a point W = ($x_0, y_0$ ) really means is that in some small neighborhood of W f is nearly linear. "Nearly" is what your definition of differentiable nails down.

Last edited: Oct 15, 2013