Is \( f \) Differentiable at \((1,0)\)?

[mahler1]

Homework Statement .

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?

 

[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 $C^1$.

 

[mahler1]

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 $C^1$.

${\frac{\partial f}{\partial y}}(1,0)$ 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.

 

[Dick]

${\frac{\partial f}{\partial y}}(1,0)$ 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.

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).

 

[brmath]

${\frac{\partial f}{\partial y}}(1,0)$ 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.

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.