I always see the example(adsbygoogle = window.adsbygoogle || []).push({});

f(x,y)={xy/(x^{2}+y^{2}) if (x,y) =/= (0,0) and 0 if (x,y)=(0,0)}

given as the example of a function where the partial derivatives exist at the origin but are not continuous there. I have a difficult time wrapping my head around this and was hoping someone could check my logic.

Now let's consider the partial derivatives f_{x}and f_{y}

f_{x}= [3x^{2}y + y^{3}]/(x^{2}+y^{2})^{2}if (x,y)=/=(0,0) and 0 if (x,y) = (0,0)

Defining it piecewise like this means f_{x}is defined at the origin, but its not continuous at the origin because of the same problem: it isn't continuous because say we approach the origin from the x axis, then this is 0, but if we approach it from the y axis, then the limit doesn't exist because you get lim x-->(0,0) 1/y = ∞. So its not continuous here because the limits don't match.

One problem is I have a difficult time understanding why the partials exist though. Because the original function f(x,y) isn't continuous at the origin. For example, if you approach the origin from the parabola y=x^{2}, you get lim (x,y)--> (0,0) [x^{3}/(x^{2}+x^{4})] (we end up using L'Hopitals Rule since this is 0/0 indeterminate, and eventually end up with the limit going to infinity because the largest power is in the denominator). So it doesn't seem like the limit of the original function exists at the origin, which would mean its not continuous at the origin, which would mean the derivatives don't exist at the origin right? Because being continuous is essential to being differentiable? Is being continuous essential to being partial differentiable?

I'm so confused!

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# Partial derivative exists at origin but not continuous there

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