I'm honestly not comfortable enough with the rules of differentiability and how they relate to partials to understand this question. Any help?
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Faires and Faires (later editions may have another author too) has one example of a function where the partials exist but the function is not differentiable. It is defined piecewise.
f(x,y) = { xy / (x^2 + y^2) if (x,y) =/= (0,0) (that's not equal)
{ 0 if (x,y) = (0,0)
f_x (0,0) = lim h->0 ( f(h,0) - f(0,0) )/ h = lim h->0 (0 - 0)/ h = 0
f_y (0,0) = lim k->0 ( f(0,k) - f(0,0) )/ k = lim k->0 (0 - 0)/ k = 0
But the function is not continuous at (0,0) and thus not differentiable.
They explain that the partial derivatives are the change in the function as you approach along a line parallel to a coordinate axis, but the derivative is the change in the function as you approach in any direction. For the function above, for any point along the line x = y, except at (0,0), f(x,y) = 1/2.
So existence of the partials is not enough.
They prove a theorem which suggests that any function that satisfies your question will be discontinuous at (1, 1). It's Theorem 14.18 on p. 800 of the 2nd edition revised.
If a function f of two variables is differentiable at (x,y), then f is continuous at (x,y).
From this it follows that if a function is discontinuous at (x,y), it can't be differentiated. The trick then is to construct a discontinuous function where the partial derivatives exist.
For your question, I'm guessing simple substitutions would suffice. Use (x-1) instead of x, (y-1) instead of y. I haven't worked through the details, but I'm reasonably confident both partials will exist. It's simply shifting the origin in a sense.
f(x,y) = { xy / (x^2 + y^2) if (x,y) =/= (0,0) (that's not equal)
{ 0 if (x,y) = (0,0)
f_x (0,0) = lim h->0 ( f(h,0) - f(0,0) )/ h = lim h->0 (0 - 0)/ h = 0
f_y (0,0) = lim k->0 ( f(0,k) - f(0,0) )/ k = lim k->0 (0 - 0)/ k = 0
But the function is not continuous at (0,0) and thus not differentiable.
They explain that the partial derivatives are the change in the function as you approach along a line parallel to a coordinate axis, but the derivative is the change in the function as you approach in any direction. For the function above, for any point along the line x = y, except at (0,0), f(x,y) = 1/2.
So existence of the partials is not enough.
They prove a theorem which suggests that any function that satisfies your question will be discontinuous at (1, 1). It's Theorem 14.18 on p. 800 of the 2nd edition revised.
If a function f of two variables is differentiable at (x,y), then f is continuous at (x,y).
From this it follows that if a function is discontinuous at (x,y), it can't be differentiated. The trick then is to construct a discontinuous function where the partial derivatives exist.
For your question, I'm guessing simple substitutions would suffice. Use (x-1) instead of x, (y-1) instead of y. I haven't worked through the details, but I'm reasonably confident both partials will exist. It's simply shifting the origin in a sense.