Why not include as a requirement that all functions must be continuous to be differentiable?

Solution 1:

Definitions tend to be minimalistic, in the sense that they don't include unnecessary/redundant information that can be derived as a consequence.

Same reason why, for example, an equilateral triangle is defined as having all sides equal, rather than having all sides and all angles equal.

Solution 2:

Because that suggests that there might be functions which are discontinuous at $a$ for which it is still true that the limit$$\lim_{t\to0}\frac{f(a+t)-f(a)}t$$exists. Besides, why add a condition that it always holds?

Solution 3:

Because then you would need to check continuity for no good reason every time you want to check for differentiability. Besides, it gives the wrong impression of being necessary to include.

Solution 4:

One possible reason is that the relationship between differentiability and continuity is more subtle in multivariable calculus.

Consider these definitions:

Let $A \subseteq \mathbb{R}^2$ and $f : A \to \mathbb{R}$ be a function. Let $a \in \operatorname{Int} A$. We define the directional derivative of $f$ at $a$ along the unit vector $v \in \mathbb{R}^2$ like this:

$$\partial_vf(a) = \lim_{h \to 0} \frac{f(a+hv) - f(a)}{h}$$

Futhermore, we say that $f$ is differentiable at $a$ if there exists a linear map $L :\mathbb{R}^2 \to \mathbb{R}$ such that

$$\lim_{h\to 0} \frac{\left|f(a+h) - f(a) - Lh\right|}{\|h\|} = 0$$ Note that $h \in \mathbb{R}^2$ here.

It can be shown that if $f$ is differentiable at $a$ then $L$ is unique and the directional derivatives exists along any unit vector $v \in \mathbb{R}^2$, being equal to $\partial_vf(a) = Lv$. Also, it implies that $f$ is continuous at $a$.

However, the converse is not true: if $f$ poseses directional derivatives along all unit vectors, $f$ does not even need to be continuous at $a$ (let alone differentiable):

Consider $f : \mathbb{R}^2 \to \mathbb{R}$ given by

$$f(x,y) = \begin{cases} 1, & \text{if $0 < y < x^2$} \\ 0, & \text{otherwise} \end{cases}$$

All directional derivatives at $(0,0)$ exist and are equal to $0$, but the function fails to be continuous at $(0,0)$.