Proving the derivative of the multivariable function $f(x,y)=\frac{x^2y}{x^2+y^2}$ exists.

Question

Given is the function $f:\mathbb{R}^2\rightarrow\mathbb{R}$ with $f(0,0)=0$ and $f(x,y)=\dfrac{x^2y}{x^2+y^2}$ for $(x,y)\neq(0,0)$.

I understand the function is continuous at $(0,0)$ and its partial derivatives exist at $(0,0)$. I need to prove the function is differentiable at $(0,0)$. The problem is that I'm unsure of what "derivative" means in this context. Is it the total derivative?

So far I thought of

$$\frac{f(x,y)-f(0,0)}{\sqrt{x^2+y^2}}=\frac{\frac{x^2y}{x^2+y^2}}{\sqrt{x^2+y^ 2}}\leq\frac{x^2y}{(x^2+y^2)y}=\frac{x^2}{x^2+y^2}\leq\frac{x^2}{x^2}=1.$$Therefore $\lim_{(x,y)\rightarrow(0,0)}\dfrac{f(x,y)-f(0,0)}{\sqrt{x^2+y^2}}\leq1$. Is this enough to prove the derivative exists in $(0,0)$?

I'd guess it means a total derivative. And no, showing that e.g. $a_n \le 1$ for every $n$ does not imply $\lim_{n\to\infty} a_n $ exists, e.g. $a_n = (-1)^n$. Same for functions. — Calvin Khor, Jul 09 '18 at 14:47
@FleetL Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work — user, Aug 06 '18 at 21:54
Possible duplicate of Proving something is not differentiable — Nosrati, Oct 07 '18 at 03:05

José Carlos Santos · Answer 1 · 2018-07-09T21:42:33.810

The derivative is a linear map $L\colon\mathbb{R}^2\longrightarrow\mathbb R$ such that$$\lim_{(x,y)\to(0,0)}\frac{\bigl|f(x,y)-f(0,0)-L(x,y)\bigr|}{\|(x,y)\|}=0.$$Since it is a linear map from $\mathbb{R}^2$ to $\mathbb R$, it is a map of the form $L(x,y)=\alpha x+\beta y$ and it can be proved that $\alpha=\frac{\partial f}{\partial x}(0,0)=0$ and that $\beta=\frac{\partial f}{\partial y}(0,0)=0$. So, if your function was differentiable at $(0,0)$, its derivative would be the null function. But that doesn't work, because this would mean that$$\lim_{(x,y)\to(0,0)}\frac{\bigl|x^2y\bigr|}{\sqrt{x^2+y^2}^3}=0,$$which is false; if $y=x>0$,$$\frac{\bigl|x^2y\bigr|}{\sqrt{x^2+y^2}^3}=\frac{x^3}{2\sqrt2x^3}=\frac1{2\sqrt2}.$$

user · Answer 2 · 2018-07-09T15:30:07.117

2

Recall that if partial derivatives exist and are continuos we are done, indeed by the "Differentiability theorem" if all the partial derivatives exist and are continuous in a neighborhood of the point then the function is differentiable at that point.

If partial derivatives are not continuos at the point we can't yet conclude anything about differentiability. We need to check directly differentiability by definition that:

$$\lim_{(h,k)\rightarrow (0,0)} \frac{\| f(h,k)-f(0,0)-(f_x(0,0),f_y(0,0))\cdot (h,k)\|}{\| (h,k)\|}=0$$

In this case since

$$f_x=\frac{2xy^3}{(x^2+y^2)^2} \quad f_y=\frac{x^2(x^2-y^2)}{(x^2+y^2)^2}$$

are not continuos at $(0,0)$, by the definition we obtain

$$\lim_{(h,k)\rightarrow (0,0)} \frac{\| f(h,k)-f(0,0)-(f_x(0,0),f_y(0,0))\cdot (h,k)\|}{\| (h,k)\|}=\frac{|h^2k|}{\sqrt{(h^2+k^2)^3}}\neq 0$$

edited Jul 09 '18 at 15:30

answered Jul 09 '18 at 14:47

user

162,563

The problem is that $f_x(0,0)$ and $f_y(0,0)$ are not even defined. – Crostul Jul 09 '18 at 14:49
As a side note, function $(f_x(0, 0), f_y(0, 0))\cdot (h, k)$ is what's called a derivative of a function of 2 variables at point $(0, 0)$ assuming differentiability – Jakobian Jul 09 '18 at 14:49
@Adam I usually call that "Directional Derivative". – user Jul 09 '18 at 14:51
Directional derivative would be in some direction, this is a linear function – Jakobian Jul 09 '18 at 14:52
@Adam Isn't it the direction given by $(h,k)$ in the expression? – user Jul 09 '18 at 14:53
1

Yes, but i meant the function of $2$ variables $(h, k) \mapsto(f_x(0, 0), f_y(0, 0))\cdot (h, k)$ – Jakobian Jul 09 '18 at 14:55
@Adam Ah ok, thanks for your clarification and useful comments! – user Jul 09 '18 at 14:56

Proving the derivative of the multivariable function $f(x,y)=\frac{x^2y}{x^2+y^2}$ exists.

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