Directional Derivatives vs. Partial Derivatives

[Buri]

I have a question about these two. I have a direction derivative at a in the direction of u defined as:

f'(a;u) = lim [t -> 0] (1/t)[f(a + tu) - f(a)]

And the partial derivative to be defined as the directional derivative in the direction of u = e_i.

My text, Analysis on Manifolds by Munkres, says that the directional derivative exists if the limit exists which I understand to mean that the right and left handed limits exists and are equal to each other. However, I asked here earlier about f(x,y) = |x| + |y| and was told the directional derivatives definition is t+ -> 0, so in fact the directional derivatives of f(x,y) do exist! But with even tweaking my definition to t+ then my partial derivatives get messed up because apparently I have to check those for t+ and t- and must equal each other.

So I'm really really confused. Can someone help me out on this? I would think that Munkres would define them correctly, so I'm probably not understanding something. Any help?

 

[HallsofIvy]

I don't know what you mean by "t+-> 0".

To find the directional derivative of f(x,y), in the direction that makes angle $\theta$ with the x-axis, you can use the unit vector $cos(\theta)\vec{i}+ sin(\theta)\vec{j}$.

In particular, for f(x,y)= |x|+ |y|, at (0, 0) we have
$$D_\theta f(0, 0)= \lim_{t\to 0}\frac{f(tcos(\theta), tsin(\theta)}{t}$$
$$= \lim_{t\to 0}\frac{|t||cos(\theta)|+ |t||sin(\theta)|}{t}= (cos(\theta)+ sin(\theta)\lim_{t\to 0}\frac{2|t||}{t}$$
and that is clearly going to give different results for t approaching 0 from above or below.

 

[Buri]

By t+ -> 0 I mean the limit from the right - so approaching through positive values.

So does this mean the limit doesn't exist then? Because I was told on here (and also checked wiki) that I'm only supposed to consider t going to 0 from positive values in directional derivatives, but for partial derivatives I must evaluate the limit for t going to 0 from positive values and negative values. I'm not exactly sure which one now? You seem to be checking left and right derivatives, which is what I initially had done. And I had the the directional deriviatves didn't exist as a result. So which is correct now? I also checked with someone else about this and they had that the directional derivatives did exist for the function we're considering, but the confusion is all a matter of definiton.

 

[HallsofIvy]

By t+ -> 0 I mean the limit from the right - so approaching through positive values.

So does this mean the limit doesn't exist then? Because I was told on here (and also checked wiki) that I'm only supposed to consider t going to 0 from positive values in directional derivatives, but for partial derivatives I must evaluate the limit for t going to 0 from positive values and negative values.

Could you give spefics? No, directional derivatives are NOT "from positive values", unless you are talking about a function, such as ln(x) which is not defined for x< 0 and so only has one sided limits.

I'm not exactly sure which one now? You seem to be checking left and right derivatives, which is what I initially had done. And I had the the directional deriviatves didn't exist as a result. So which is correct now? I also checked with someone else about this and they had that the directional derivatives did exist for the function we're considering, but the confusion is all a matter of definiton.

 

[Buri]

Well the example you considered:

f(x,y) = |x| + |y|

I had initially thought that the directional derivatives did not exist since the left and right limits weren't equal. But only taking the limit for t > 0 then they do. A user on here said they did exist (he was taking t through positive values). So I got all confused.

So I guess we're supposed to alter the definition depending on the function we're trying to find the directional derivatives for?

 

[HallsofIvy]

Perhaps you are thinking of this case:
Approaching (x, y) from the direction that makes angle $\theta$ with the x- axis,
$$D_\theta(f)= \lim_{r\to 0}\frac{f(x+ rcos(\theta), y+ rsin(\theta))- f(x, y)}{r}$$

There r goes to 0 from above because r, a distance, is always positive. But as long as the function is differentiable at (x, y), that will give the same result, with reversed sign, as approaching along angle $2\pi+ \theta$ which is exactly the same as reversing the sign on r.

 

[Buri]

Sorry its been a while, I've been rather busy with midterms. So I suppose also (contraspositive) that if not all directional derivatives exist at a point then the function is not differentiable at the point right?

Btw, geometrically what is the directional derivative of a function f: R -> R? Is it like the slope of a messed up tangnet? That's what it seems like..

 

[Landau]

It's probably a matter of definition whether to take the limit from both sides or just one side, but I don't think the latter is standard. Could you post a link to that thread you referred to?

And you are right, if f is differentiable at some point (by which I mean the 'total derivative' exists at that point), then the dir. der. at that point in all directions exist. However, the converse does not hold: if at some point f is directionally differentiable in all directions, it need not follow that f is differentiable at that point. An example:

$$f:\mathbb{R}^2\to \mathbb{R}^2$$
$$f(x,y):=\begin{cases}\frac{xy^2}{x^2+y^4}\text{ if }(x,y)\neq (0,0) \\ 0\text{ if }(x,y)= (0,0) \end{cases}.$$
Exercise: check my statement, in particular show that the dir. der. of f at (0,0) in every direction exists, while f is not differentiable at (0,0).

@your last point: the dir.der. of a function f:R\to R is the same as the ordinary derivative, since R has only one 'direction': it's a 1-dimensional vector space.

 

[Buri]

It's probably a matter of definition whether to take the limit from both sides or just one side, but I don't think the latter is standard. Could you post a link to that thread you referred to?

Here they are:

https://www.physicsforums.com/showthread.php?t=436047

https://www.physicsforums.com/showthread.php?t=435468

@your last point: the dir.der. of a function f:R\to R is the same as the ordinary derivative, since R has only one 'direction': it's a 1-dimensional vector space.

But I could take scalar multiples of the directional vector, couldn't I? And this would distort it a bit, right?

Also I notice everyone whose helped me uses polar coordinates or does directional derivatives with angles. Why is this? See my definition of a directional derivative at a in the direction of u is:

D_u f(a) = lim[t → 0] 1/t[f(a + tu) - f(a)]

On a problem set I handed in a while back (and got back already) I had to show the following function was not differentiable at 0 but all directional derivatives exist at 0:

Let f(x,y) = √|xy| if x ≥ 0 OR -√|xy| if x < 0

So if I let v = (h,k) I have:

D_v f(a) = lim[t → 0] 1/t[f(t(h,k) - f(a)]

But now it gets a bit confusing, because of all the cases.

For h ≥ 0 I have:

D_v f(a) = lim[t → 0] √|(th)(tk)|/t = lim[t → 0] |t|√|hk|/t

See, but how do I really know that f(t(h,k)) = √|(th)(tk)| ? Because when I take the left and right limits t > 0 or t < 0 so it would change the value of the function since the first coordinate could turn out to be greater than zero and less than zero. So would I end up with 4 cases instead? Like below:

Case 1: t > 0 and h ≥ 0

D_v f(a) = lim[t → 0] √|(th)(tk)|/t = lim[t → 0] |t|√|hk|/t = lim[t → 0] t√|hk|/t = |hk|

Case 2: t < 0 and h ≥ 0

D_v f(a) = lim[t → 0] -√|(th)(tk)|/t = lim[t → 0] -|t|√|hk|/t = lim[t → 0] -(-t)√|hk|/t = √|hk|

So for h ≥ 0 the limit exists.

Case 3: t > 0 and h < 0

D_v f(a) = lim[t → 0] -√|(th)(tk)|/t = lim[t → 0] -|t|√|hk|/t = lim[t → 0] -t√|hk|/t = -√|hk|

Case 4: t < 0 and h < 0

D_v f(a) = lim[t → 0]√|(th)(tk)|/t = lim[t → 0]|t|√|hk|/t = lim[t → 0] -t√|hk|/t = -√|hk|

So the limit exists for h < 0.

However, the right and left limits don't equal each other. So the limit shouldn't exist right? But apparently they do. My TA happened to mark this question and I sneakily (I lost a mark on it because of it lol) removed the negative on the last one so the left and right should be equal, but then he inserted the ± into it. But wouldn't that mean the limit doesn't exist? I asked him about it, but he didn't seem to understand my confusion. So help would be amazing.

 

[Buri]

Wait a second. They do exist. I've broken up the cases for v = (h,k) for h ≥ 0 and h < 0, but when I do the limits they (the cases for h ≥ 0 and h < 0) don't need to be equal since these aren't the left and right limits. So D_v f(a) = ±√|hk|, that's why when my TA inserted the ± it wasn't contradicting that the directional derivatives indeed do exist. I'm correct, right?

 

[Landau]

Wait a second. They do exist. I've broken up the cases for v = (h,k) for h ≥ 0 and h < 0, but when I do the limits they (the cases for h ≥ 0 and h < 0) don't need to be equal since these aren't the left and right limits. So D_v f(a) = ±√|hk|, that's why when my TA inserted the ± it wasn't contradicting that the directional derivatives indeed do exist. I'm correct, right?

Yes, of course. For each FIXED direction v=(h,k) you have to compute the limit. The case h>=0 and h<0 give you different directions, totally independent.

By the way, I see why some would only talk about the limit for t>0: we are taking the derivative at a in the direction of v, so informally "we start at a and move infinitesimally towards v". If t<0 then "we start at a and move towards -v", so to speak.

In one of the threads you linked to, I have explained some things about the total derivative and the directional derivative. Please have a look :)