Question about limit definition of partial derivative

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The discussion centers on the definitions of the partial derivative, specifically comparing two limit definitions: one using the general point (x, y) and the other using a specific point (x0, y0). The function in question is defined piecewise, with a specific focus on the point (0, 0). The second definition simplifies the evaluation of the derivative at this point, making it easier to demonstrate differentiability. The user seeks clarification on which definition is conventionally accepted in mathematical practice.

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I've seen it written two different ways:

$$\frac{\partial f}{\partial x} = \lim\limits_{h \rightarrow 0} \frac{f(x + h, y) - f(x,y)}{h}$$

and

$$\frac{\partial f}{\partial x} = \lim\limits_{h \rightarrow 0} \frac{f(x_0 + h, y_0) - f(x_0,y_0)}{h}$$

where the latter evaluates the function at the respective point before plugging it into the definition of the limit. For example, the function ##f(x,y) = \begin{cases}
\frac{x^2 y^4}{x^4 + 6y^8}, & \text{if }(x,y) \neq (0,0) \\
0, & \text{if }(x,y) = (0,0)
\end{cases}##

I want to determine if ##\frac{\partial f}{\partial x}## is differentiable at ##(0,0)##.

Using the second limit definition would make showing the existence of ##\frac{\partial f}{\partial x}## considerably easier, since ##y_0## makes the first term in the limit ##0##, and ##f(x_0,y_0)## is defined to be ##0##.

But using the first definition, we have to evaluate:

$$\frac{(x+h)^2 y^4}{(x+h)^4 + 6y^8} - \frac{x^2 y^4}{x^4 + 6y^8}$$

I'm hoping the "real" or at least usable definition is the second one, but which one is the one we're supposed to use in practice to be technically/mathematically correct?
 
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