I am told that a function $f:\mathbb{R}^{n}\to \mathbb{R}$ is differentiable at a point $\mathbf{x}=(x_1,x_2,…,x_n)$ if $\Delta f$ is of form
$$\sum_{i}^{n}\frac{\partial f}{\partial x_i}\Delta x_i+\sum_{i}^{n} \epsilon_i \Delta x_i$$
Where
$$\lim_{\mathbf{\Delta x}\to \mathbf{0}}\epsilon_{i}=0$$ for every $1\leq i\leq n$.
It seems arduous to me to use such a definition. Why don't mathematicians define differentiability by the existence of partial derivatives, which seems natural?
Now, I am aware that I can define things how I want to, but my goal is to understand why it is commonly defined this way. Are there some underlying properties that make this definition superior?
Best Answer
The reason for this is because a point $P$ in the domain could be reached by more than one path.
We'll try to understand this using a function of two variables since it is very easy to visualise and get a good intuition.
For a function of two variables, the two partial derivatives are arrived at by intersecting the surface of the graph with vertical planes parallel to $xz$- and $yz$-planes, creating a curve in each plane called a trace at $P$.
The partial derivatives are seen as the slope of tangent to these curves at the point $P$. Now it would be natural to expect the following: if you were to turn any of those vertical planes slightly about $P$, keeping it vertical, but no longer parallel to the coordinate plane, the tangent to the new trace at $P$ slightly different to what it was before. Right?
Obviously, this is not guaranteed by the existence of just the two partial derivatives. For the function to be differentiable, it should not change abruptly as you move slightly around a point's neighborhood. Hence that definition.
Makes sense?