Some clarification needed on the Relation between Total Derivative and Directional Derivative

I will consider here functions of several variables only.

If both directional derivative $D_{v}f(x)$ at $x$ along $v$ and total derivative $D f(x)$ at $x$ exist then $$D_{v}f(x)=Df(x)(v).$$ Existence of total derivative ensures that of directional derivative in every direction but not the other way round.

There are functions who have, at some point of the domain, directional derivative in every direction but not differentiable at that point i.e. the total derivative at that point does exist.

Now, all of my knowledge is theoretical. I cannot see the picture clearly, i.e. the picture of the two kinds of derivative existing together, or one existing and not the other - how do these work?

I mean some geometrical interpretation for say $2$ or $3$ dimensional space would help.

I am so confused with this thing, I am not even sure if I have managed to convey my problem properly.

Please help with some clarification.

Thanks.

Solution 1:

There’s a very nice discussion of this topic on Math Insight that might be helpful. The key point for your question is that directional derivatives only “look” along straight lines, but the total derivative (also called the differential) requires you to look at all ways to approach the point. For a function on the real line, there are only two ways to do this—from the right and left—but once you move to functions on the plane and beyond, there are suddenly many, many paths available.

For a two-dimensional surface in three dimensions, you can think of the differential as specifying the tangent plane to the surface at a point. For the differential to exist, the tangent vectors to the surface at this point of all smooth paths along the surface which pass through the point must lie in the same plane. Directional derivatives correspond to only those paths whose projections onto the $x$-$y$ plane are straight lines. There are clearly many other smooth paths through the point.

Solution 2:

How about this? The function $$f(x,y) = \begin{cases} 1 \text{ if $y = x^2$ and $x \neq 0$} \\ 0 \text{ otherwise}\end{cases}$$

Draw a picture. You will see that every directional derivative exists and is equal to $0$, but the $x-y$ plane is not a good approximation for the function $f$ near $(0,0)$.