# Directional derivative of a function and continuity

I am studying directional derivatives, and I am stuck with how to visualize them. What it means geometrically to take the derivative of a function f at a point a relative to the vector u?

My own understanding is that if i am given a point, I can draw infinitely many tangent lines, but when there is a vector in accordance to which I should take the derivative, there is only one tangent line.

And there is a further question related to this. The book notes that directional derivative does not imply that the function is continuous, but does not elaborate. What does that mean?

Well, here's a simple way of thinking about it. Think about the function $f$ as a scalar function (a field', in physics jargon) in $R^3$, i.e. $f=f(x,y,z)$. In general the function does not have a constant value, i.e. it varies from point to point in space.

Let's say the function $f$ is 0 at the origin. You can now move off from the origin in any direction (towards (1,1,1) say) and see how the value of your function changes. Or you can move along one of the axes (towards (1,0,0) or (0,1,0)). You don't have to travel very far (just infinitesimally) to sense changes in the value of the function, but you have the choice of a full sphere's worth of directions to explore.

The important thing is that because the function has different values at different locations in space, the rate of change of the function will also differ (in general) depending on which direction you set off in from the origin.

Clearly, there was nothing special about the origin in the argument above, and you could start at any point in the domain of the function and move outward from there, and you'd see different rates of change in different directions.

Thus, you have the concept of the rate of change of a (scalar) function $f$ at a given point in a particular direction, which is precisely what the directional derivative measures.

We can now construct a limit' type definition for the directional derivative based on the above, i.e. if you think of the function $f$ as taking in a position vector $\overline{p}=(x,y,z)$ and spewing out a number, you'd have the directional derivative of the function $f$ at the point $\overline{p}$ in the direction $\hat{n}$ given by

$$\nabla_{\hat{n}}(f)(\overline{p}) = \lim_{h \rightarrow 0} \frac{f(\overline{p}+h\hat{n}) - f(\overline{p})}{h}$$

Appended a correction: It is true that the function need not be continuous at a point before you take its directional derivative there. However, this does not mean the function is differentiable at a point where it is not continuous (just as in single variable calculus).