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 $\frac{\partial f}{\partial x}$ exists 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?
Best Answer
$y_0$ does not mean "set $y$ to $0$", it just indicates a particular value of $y$. The two formulas suggest finding the partial derivative for any general point $(x,y)$ and finding the partial derivative for a specific point $(x_0,y_0)$ [but not necessarily $(0,0)$]. Of course that's just interpretation though.