Hi,(adsbygoogle = window.adsbygoogle || []).push({});

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

[tex]

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

[/tex]

have total differential in [0,0]?

First I computed partial derivatives:

[tex]

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

[/tex]

[tex]

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

[/tex]

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:

[tex]

\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

[/tex]

[tex]

\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

[/tex]

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

[tex]

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

[/tex]

for any

[tex]

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

[/tex]

and must satisfy the limit

[tex]

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

[/tex]

I can write it this way:

[tex]

\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}}

[/tex]

When I put

[tex]

h_2 = kh_1

[/tex]

I can write

[tex]

\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

[/tex]

And thus I say thatfdoesn't have total differential in [0,0].

Is this correct approach?

Thank you for checking this out.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Is this ok?

**Physics Forums | Science Articles, Homework Help, Discussion**