Question

Show that f differentiable at (0,0) ?

Consider the function :
$$f: \mathbb{R}^2 \rightarrow \mathbb{R} , (x,y) \mapsto
\begin{cases}
0 & \text{for } xy=0 \
{x^2y^2cos}
\frac{1}{x^2y^2} & \text{for } xy \neq 0
\end{cases} $$

Show that f differentiable at (0,0) .

I don't know how to tackle this problem. Can someone give some hints or solve the problem? Thanks :)

Answer 1

Hint:
You need to show that `lim_((x,y)rarr(0,0)) x^2y^2cos(1/(x^2y^2))=0`

Answer 2

See below.

Explanation:

According with Gateau's differentiation

`F(x,y)` is differentiable at `x_0,y_0` if there exists

`lim_(epsilon->0) (F(x_0+epsilon h_1,y_0+epsilon h_2)-F(x_0,y_0))/epsilon`

In our case

`(x_0,y_0) = (0,0)` so

`lim_(epsilon->0) (F(epsilon h_1, epsilon h_2)-F(0,0))/epsilon = lim_(epsilon->0)epsilon^2h_1^2h_2^2 cos(1/(epsilon^2 h_1^2 h_2^2))/epsilon = 0`

so according with Gateau's differentiation definition, `F(x,y)` is Gateau differentiable at `(0,0)`

There are other differentiation criteria like Lipchitz's or Fréchet's. Of those, the strongest differentiation method is Fréchet's.

The next step is to show Frechet's differentiability.