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I was asked to find the partial derivatives at every point for the following function:

$$ f(x,y)= \left\{ \begin{array}{c} \frac{(x^2+y)^2}{x^4+y^2} \qquad (x,y)\neq(0,0) \\ \;\;\; 1 \qquad \quad (x,y)=(0,0) \\ \end{array} \right. $$

Now, for all $(x,y)\neq(0,0)$ (after some legwork) the answers are: $$\frac{\partial f(x,y)}{\partial x}=\frac{4yx(y^2-x^4)}{(x^4+y^2)^2}$$ $$\frac{\partial f(x,y)}{\partial y}=\frac{2x^2(x^4-y^2)}{(x^4+y^2)^2}$$

My question is about when $(x,y)=(0,0)$. Is $f$ discontinuous there? What can I say about the partial derivatives at that point? They don't exist there? They exist but are not differentiable there?

Thank you.

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  • $\begingroup$ How did you get discontinuity? $\endgroup$
    – Juanito
    May 18, 2017 at 22:05
  • $\begingroup$ Sorry! Correcting it. $\endgroup$
    – Friedman
    May 18, 2017 at 22:10

2 Answers 2

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The function is indeed discontinuous at $(0,0)$, because $\lim_{t\to 0}f(x(t),y(t))=0$ for $x(t)=t,\ y(t)=-t^2$.

In order to find $\frac{\partial f}{\partial x}(0,0)$ you must evaluate, as per definition $$\lim_{t\to0}\frac{f(t,0)-f(0,0)}t=\lim_{t\to0}\frac{\frac{(t^2+0)^2}{t^4+0}-1}{t}$$ and similarly for $\frac{\partial f}{\partial y}(0,0)$.

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For $y_2\ne 0$ and $y_1=0$ we have $(f(0,y_2)-f(0,y_1))/(y_2-y_1)=0.$ So $\partial f(x,y)/\partial y=0$ when $(x,y)= (0,0).$ Similarly $\partial f(x,y)/\partial x=0$ when $(x,y)=(0,0).$

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