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

I have this excercise to do:

Let's have functionfdefined as follows:

[tex]

f(x,y) = \frac{xy(x^2 - y^2)}{x^2 + y^2},\ \ [x,y] \neq [0,0]

[/tex]

[tex]

f(x,y) = 0,\ \ [x,y] = [0,0]

[/tex]

Find out, whether this function has total differential in [0,0].

Well, first I observed that partial derivatives aren't continuous in [0,0], which is what I expected.

So I computed them from the definition:

[tex]

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

[/tex]

[tex]

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

[/tex]

So, if the total differential exists in [0,0], it must be zero linear transform.

Now I have to checkout, if such a transform satisfies the limit and thus it is total differential:

[tex]

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

[/tex]

If I rewrite it in a slightly different way ( I put [itex]h = (h_1,h_2)[/itex] and [itex]h_2 = kh_1[/itex], I get

[tex]

\lim_{h_1 \rightarrow 0} \frac{\frac{kh_1^2(h_1^2 - k^2h_1^2)}{h_1^2(1+k^2}}{h_1\sqrt{1+k^2}} = \frac{k(1-k^2)}{\sqrt{(1+k^2)^3}} \lim_{h_1 \rightarrow 0} h_1

[/tex]

Well, I don't know whether this goes to zero so I tried another way, ie. approaching [0,0] on the x-axis and thus putting [itex]h_2 = 0[/itex]

The limit I got was zero, so I can't say at this moment that the total differential doesn't exist in [0,0]. Anyway, according to right results our professor gave us, total differential doesn't exist in [0,0].

How could I prove that?

Thank you very much.

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# Homework Help: Total differential question

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