Differentiability in two variables - directional derivative & gradient

I have read a chapter about differentiability in two variables. I now have two questions:

1. Why do we need the constraint that $|\vec{u}|=1$ when we calculate the directional derivative?
2. Definition of gradient: $\nabla f = \frac{\partial f}{\partial x_1 }\mathbf{e}_1 + \cdots + \frac{\partial f}{\partial x_n }\mathbf{e}_n$ but why do we need f to be differetiable? Is it simply becuase we "take the partial derivatives"? I have learned the pure defn of differetiable, but maybe I don't have the "gut feeling" of what it really is, in simple language.

1) You are free to choose $u$ not unitary. But you have to be careful. For example, if $|u|=2$ you get double the amount you get with $|u|=1$, as it means that you are "moving twice as fast".

For example, consider a function $f(x,y)$, and consider the "curve" $(x=t, y=0)$, which is a straight line along the $x$ axis.

How do we "see" $f$ from "inside" the curve? We get: $$f(x(t), y(t)) = f(t,0).$$

The derivative of $f(t,0)$ with respect to $t$ is exactly the directional derivative in the direction of $x$.

Now pick the "curve" $(x=2t, y=0)$, which is a straight line like before, but now the "speed" is double. What happens now if you calculate: $$f(x(t), y(t)) = f(2t,0),$$

and derive it with respect to $t$?

2) "Differentiable" intuitively means that "its partial derivatives exist and behave well". Which translates to: "if $f$ is not differentiable, it makes no sense to talk about its gradient."

• can you elaborate on 1) ? – jacob Apr 19 '14 at 15:13
• @jacob Edited the question, adding an example...Let me know if it is clear! – geodude Apr 19 '14 at 15:22