7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Do you know the complete definition of total derivative i.e. it is a linear map and all that? $\endgroup$ – Landon Carter Sep 8 '15 at 16:14
  • $\begingroup$ @LandonCarter : Yes I have read all that in my book but like I said , the picture is not clear to me . $\endgroup$ – user118494 Sep 8 '15 at 16:25
5
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $(0,0)$.

$\endgroup$
1
$\begingroup$

We had a conversation before about both of us studying multivariable calculus before. It seems that we are in similar chapters. So I want to help you if I can. First of all, we have to clarify your first statement. You say that if both directional derivative $D_{v}f(x)$ at $x$ along $v$ and total derivative $Df(x)$ at $x$ exist then

$$D_{v}f(x)=Df(x)(v)$$

This is not always the case. Sometimes both total derivative and directional derivative may exist, but

$$D_{v}f(x) \neq Df(x)(v)$$

For example, consider the function

$$f(x,y) = \begin{cases} \frac{x^2y}{x^2+y^2} \text{ if $(x,y) \neq (0,0)$} \\ 0 \text{$\qquad$ if $(x,y) = (0,0)$}\end{cases}$$

For this function, $D_vf(x)=1/2$ for $x=(0,0)$ and $v=\langle 1,1 \rangle$. However, $Df(x)v=0$. As you see, knowing only total derivative or directional derivative is not helpful. You should not expect a relationship between them unless required criterions are met.

They are related ($D_{v}f(x)=Df(x)(v)$) if $f$ is differentiable at $x$. So you need a differentiability criterion. The following theorem helps: If, for a function $f:U \to \mathbb R^m$, all partial derivatives exist and they are continous on $U$, then $f$ is differentiable on $U$.

$\endgroup$
  • $\begingroup$ Are you sure about that? $f(h\cos\theta,h\sin\theta) = h\cos^2\theta\sin\theta$, so all directional derivatives exist and are equal to zero. $\endgroup$ – amd Jun 22 '18 at 23:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.