Checking Total Differential of f(x,y) in [0,0]

[twoflower]

Hi,

I started computing excercises on total differential and I would like to know if I'm doing it correctly. Could you please check it? Here it is:

Does the function

$$f(x,y) = \sqrt[3]{x^3+y^3}$$

have total differential in [0,0]?

First I computed partial derivatives:

$$\frac{\partial f}{\partial x} = \frac{x^3}{\sqrt[3]{(x^3 + y^3)^2}}$$

$$\frac{\partial f}{\partial y} = \frac{y^3}{\sqrt[3]{(x^3 + y^3)^2}}$$

I see that partial derivatives are continuous everywhere with the exception of the point [0,0].

For the point [0,0] I have to compute partial derivatives from definition using the limit:

$$\frac{\partial f}{\partial x}(0,0) = \lim_{t \rightarrow 0} \frac{f(t,0) - f(0,0)}{t} = \lim_{t \rightarrow 0} \frac{t}{t} = 1$$

$$\frac{\partial f}{\partial y}(0,0) = \lim_{t \rightarrow 0} \frac{f(0,t) - f(0,0)}{t} = \lim_{t \rightarrow 0} \frac{t}{t} = 1$$

So in the case that total differential in the point [0,0] exists, it must be of form:

$$L(h) = \frac{\partial f}{\partial x}(0,0) h_1 + \frac{\partial f}{\partial y}(0,0) h_2 = h_1 + h_2$$

for any

$$h = (h_1, h_2) \in \mathbb{R}^2$$

and must satisfy the limit

$$\lim_{||h|| \rightarrow 0} \frac{f(0,0) + h) - f(0,0) - L(h)}{||h||} = 0$$

I can write it this way:

$$\lim_{[h_1,h_2] \rightarrow [0,0]} \frac{\sqrt[3]{h_1^3 + h_2^3} - 0 - h_1 - h_2}{\sqrt{h_1^2 + h_2^2}}$$

When I put

$$h_2 = kh_1$$

I can write

$$\lim_{h_1 \rightarrow 0} \frac{ \sqrt[3]{h_1^3 + k^3h_1^3} - h_1kh_1}{\sqrt{h_1^2 + k^2h_1^2}} = \frac{\sqrt[3]{1+k^3} - 1 - k}{\sqrt{1 + k^2}} \neq 0$$

And thus I say that f doesn't have total differential in [0,0].

Is this correct approach?

Thank you for checking this out.

 

[HallsofIvy]

You have the partial derivatives wrong- though it may just be a typo.
$$f_x(x,y)= \frac{x^2}{^3\sqrt{(x^3+ y^3)^2}}$$

 

[twoflower]

You have the partial derivatives wrong- though it may just be a typo.
$$f_x(x,y)= \frac{x^2}{^3\sqrt{(x^3+ y^3)^2}}$$

You're right, it's a typo. Is it ok otherwise?