A Condition For Differentiability

Consider the function \(f(x)\) at point \(x=a\). If \(y=f(x)\) and \(x\) changes from \(a\) to \(a + \Delta x\), we define the increment of \(y\) as:

$$ \Delta y = f(a + \Delta x) - f(a) $$

According to the definition of a derivative, we have:

$$ \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = f'(a) $$

Let \(\epsilon\) be the difference between these two:

$$ \epsilon = \frac{\Delta y}{\Delta x} - f'(a) $$

If we define \(\epsilon\) to be 0 when \(\Delta x = 0\), then \(\epsilon\) becomes a continuous function of \(\Delta x\). Thus, for a differentiable function \(f\), we can rearrange and write:

$$\begin{gather} \Delta y = f'(a) \Delta x + \epsilon \Delta x \\ \text{where } \epsilon \to 0 \text{ as } \Delta x \to 0 \end{gather}$$

Now consider a function of two variables, \(z=f(x,y)\), and suppose \(x\) changes from \(a\) to \(a+ \Delta x\) and \(y\) changes from \(b\) to \(b + \Delta y\), then the corresponding increment of \(z\) is:

$$\Delta z = f(a + \Delta x, b + \Delta y) $$

If \(f\) is differentiable at \((a,b)\), then by analogy of the function with one variable, the following should hold true:

$$\begin{gather} \Delta z = f_x(a,b) \Delta x + f_y(a,b) \Delta y + \epsilon_1 \Delta x + \epsilon_2 \Delta y \\ \text{where } \epsilon_1 \text{ and } \epsilon_2 \to 0 \text{ as } (\Delta x, \Delta y) \to (0, 0) \end{gather}$$

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