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How do we show that a function is continuously partially differentiable?

For example if we consider the function $$f(x, y) =\begin{cases} xy\frac{x^2-y^2}{x^2+y^2}& \text{ if } (x, y) \neq (0.0) \\ 0 & \text{ if } (x, y) =(0, 0)\end{cases}$$ to show that it is partially differentiable do we just calculate the partial derivatives to show that these exist?

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EDIT:

To show that the partialderivative as for$x$ is continuous I have done the following:

\begin{align*}\left |\frac{\partial{f}}{\partial{x}}(x,y)\right |&=\left |\frac{y(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2}\right |\\ & \leq \left |\frac{y(x^4+4x^2y^2+y^4)}{(x^2+y^2)^2}\right |\\ & =\left |\frac{y(x^4+2x^2y^2+y^4+2x^2y^2)}{(x^2+y^2)^2}\right |\\ & =\left |\frac{y((x^2+y^2)^2+2x^2y^2)}{(x^2+y^2)^2}\right |\\ & =\left |\frac{y((x^2+y^2)^2)}{(x^2+y^2)^2}+\frac{y(2x^2y^2)}{(x^2+y^2)^2}\right |\\ & =\left |y+\frac{y(2x^2y^2)}{(x^2+y^2)^2}\right |\end{align*}

How can we continue?

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    $\begingroup$ You need to show partial derivative exists and the calculated partial derivative must be continuous. $\endgroup$
    – Balaji sb
    May 6 at 4:36
  • $\begingroup$ Could you please take a look at the edit part of my post above? How do we continue to show the continuity? @Balajisb $\endgroup$
    – Mary Star
    May 6 at 8:55
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    $\begingroup$ Rational functions in (x,y) is continuous unless denominator becomes $0$. For rational functions: if u check only at the points: a) denominator of partial derivative = 0 and b) Special points like (0,0) then u r fine. $\endgroup$
    – Balaji sb
    May 6 at 9:08
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    $\begingroup$ To prove that rational functions in $(x,y)$ is continuous at points other than poles (denominator=$0$) use/prove that ratio ,product,sum of continuous function are continuous. The proof of these facts are available in Rudin. $\endgroup$
    – Balaji sb
    May 6 at 9:10
  • $\begingroup$ Ok! But in this case we have to check that the partialderivatives are continuous in $(0,0)$ or not? So do we check if the limit of the function in (x,y) when $(x,t)\to(0,0)$ goes to $0$ ? @Balajisb $\endgroup$
    – Mary Star
    May 6 at 9:59
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Note that $$f'_x(0) = \lim_{x\to 0}xy\frac{x^2 - y^2}{x^2 + y^2} = 0$$ $$f'_y(0) = \lim_{y\to 0}xy\frac{x^2 - y^2}{x^2 + y^2} = 0$$

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  • $\begingroup$ So do we need to show that the partial derivatives exist in (0,0)? $\endgroup$
    – Mary Star
    May 6 at 5:44
  • $\begingroup$ Could you please take a look at the edit part of my post above? How do we continue to show the continuity? $\endgroup$
    – Mary Star
    May 6 at 8:55

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