Undergrad Differentiability of a Multivariable function

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The discussion centers on the differentiability of the multivariable function f(x,y) = y^3/(x^2+y^2) at the origin. It is noted that while the function is undefined at (0,0), defining f(0,0) = 0 allows for the use of the limit definition of a partial derivative. The geometric interpretation of the partial derivative as the intersection of a plane with the surface supports this approach. However, if f were defined as a different value at the origin, such as 2, the limit definition would not apply. The conversation highlights that partial derivatives can exist even for functions that are not continuous at a point.
lys04
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I’m having a little confusion about part b of this question as to why I am allowed to use the limit definition of a partial derivative.
Here’s what I think:
I know that y^3/(x^2+y^2) is undefined at the origin but it does approach 0 when it GETS CLOSE to the origin. So technically defining f(x,y)=0 fills in that hole and it becomes a smooth curve and so I can use the limit definition? (Because the geometric interpretation of a partial derivative, at least with respect to x, is the intersection of y=y_0 with the surface, which becomes a 2d curve, and then I take the derivative wrt x.)
If instead f is defined to be some other number like 2 at the origin then this will not work?
 

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The partial derivative at ##(0,0)## is defined as:
$$\frac{\partial f}{\partial x} = \lim_{h \to 0}\frac{f(h, 0) - f(0, 0)}{h}$$PS I thought partial derivatives were only defined for a continuous function, but apparently not!
 
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There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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