# Definition of differentiability of function of two variables

I assume that the reader may be well versed with the various notations of partial derivatives.

My book gives the following definition of differentiability:

If $$\Delta f(x,y)$$ can be expressed in the form:

$$\Delta f(x,y)=f_x(a,b)\ \Delta x + f_y(a,b)\ \Delta y + \varepsilon_1\ \Delta x + \varepsilon_2\ \Delta y$$;

where $$\varepsilon_1 \rightarrow 0$$ and $$\varepsilon_2 \rightarrow 0$$ as $$(\Delta x,\Delta y) \rightarrow (0,0)$$;

then $$f(x,y)$$ is differentiable at $$(a,b)$$

However my calculation gives:

Provided that $$f_x(a,b)$$, $$f_y(a,b)$$ and $$f_{xy} (a,b)$$ exist,

$$\Delta f(x,y)=f_{xy}(a,b)\ \Delta x \Delta y+f_x(a,b)\ \Delta x + f_y(a,b)\ \Delta y + \varepsilon_1\ \Delta x + \varepsilon_2\ \Delta y + \varepsilon_3\ \Delta x \Delta y$$

where $$\varepsilon_1 \rightarrow 0$$, $$\varepsilon_2 \rightarrow 0$$ and $$\varepsilon_3 \rightarrow 0$$ as $$(\Delta x,\Delta y) \rightarrow (0,0)$$

Does this mean if $$f_x(x,y)$$ is not a function of $$y$$ , i.e. $$f_{xy}(x,y)=0$$ i.e. equation $$(1)$$ is obeyed and therefore $$f(x,y)$$ is differentiable at $$(a,b)$$???

In other words, if $$f_x(x,y)$$ is not a function of $$y$$, then is $$f(x,y)$$ is differentiable at $$(a,b)$$???

The calculation is simple, trivial but a bit lengthy. If the reader feels anything wrong, I can present the calculation

As stated in the question, a definition of differentiability is

$$\Delta f(x,y)=f_x(a,b)\ \Delta x + f_y(a,b)\ \Delta y + \varepsilon_1\ \Delta x + \varepsilon_2\ \Delta y \tag{1}\label{eq1}$$

I'm not quite sure how you came up with your calculation of

$$\Delta f(x,y) = f_{xy}(a,b)\ \Delta x \Delta y+f_x(a,b)\ \Delta x + f_y(a,b)\ \Delta y + \varepsilon_1\ \Delta x + \varepsilon_2\ \Delta y + \varepsilon_3\ \Delta x \Delta y \tag{2}\label{eq2}$$

For one thing, why did you choose just $$f_{xy}(a,b)$$ and not use $$f_{y\, x}(a,b)$$ as well? Regardless, the $$2$$ equations are not actually inconsistent in the limiting case as $$(\Delta x,\Delta y) \rightarrow (0,0)$$. This is because the extra $$2$$ terms you have are both bounded values multiplied by the second order delta values of $$\Delta x \Delta y$$ while all of the original values on the right are a multiple of just $$\Delta x$$ or $$\Delta y$$.

A related question about this was asked here at First principles derivation of area under a curve giving rise to an unexpected term before taking limits. As the comment there by Andy Walls basically states, when $$\Delta x$$ and $$\Delta y$$ become small, then their product becomes extremely small, e.g., $$0.0000001 \times 0.0000001 = 0.000000000001$$.

To help show why this works, consider that $$\Delta x$$ and $$\Delta y$$ are changing proportionally to each other, i.e., that $$\frac{\Delta x}{\Delta y} = k$$, for some non-zero constant $$k$$, so $$\Delta x = k\varepsilon_4$$ and $$\Delta y = \varepsilon_4$$ for some small real $$\varepsilon_4$$. Substitute this into the RHS of \eqref{eq2} and divide both sides by $$\varepsilon_4$$ to get

$$\frac{\Delta f(x,y)}{\varepsilon_4} = f_{xy}(a,b)\ k\varepsilon_4 + f_x(a,b) k + f_y(a,b) + \varepsilon_1\ k + \varepsilon_2\ + \varepsilon_3\ k\varepsilon_4 \tag{3}\label{eq3}$$

Now, taking the limit as the various $$\varepsilon_i \to 0$$ gives you on the RHS

$$f_x(a,b) k + f_y(a,b) \tag{4}\label{eq4}$$

Note doing the exact same calculations in \eqref{eq1} gives you the same result. As you can see, when you're taking limits going to $$0$$, only the lowest order terms will survive.

I hope this answers your question well enough. Keep in mind I used just a somewhat restricted case of $$\Delta x \to 0$$ and $$\Delta y \to 0$$ because I thought it'd be more straight forward & simpler than showing the general case, but you may wish to try that yourself to confirm your understanding of this issue.