# Talk:Directional derivative

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## Directional derivatives along normalized vectors only?

should it be specified that v has to be a normalized vector? --anon

Not necessarily. You can also take the directional derivative in the direction of the zero vector, as no division by zero is involved. Oleg Alexandrov (talk) 03:08, 9 March 2006 (UTC)
Hmmmm...It seems pointless to allow the directional derivative to not be normalized. mathworld seems to specify that the direction ought to be normalized too. What would be the purpose of allowing the directional derivative in the zero vector direction? --anon

One has

$D_{\vec{v}}{f}= \nabla f \cdot \vec{v}$

where $\nabla$ is the gradient.

I see no reason to require that you must do dot products only with unit vectors in the formula above. Oleg Alexandrov (talk) 03:08, 10 March 2006 (UTC)

We aren't talking about ONLY doing dot products with unit vectors, but that the directional derivitive is definied as a gradient of a function dotted with the unit vector of the vector in question evaluated at a point. So long as the unbit point is made, the equation is fine. I can tell that the data in this article came from planet math, which like wikipedia is user editited. Even the the eratta state that the Vector in question is unitary. However, don't take my word for it;
* The directional derivative at Wolfram
* Multivariate Calculus at usd.edu
* The directional derivative at lamar.edu
-- Dbroadwell 19:59, 3 May 2006 (UTC)

Well, if you say "derivative along a vector", that vector does not need to be of length one. If you say "derivative along a direction", then yes, a direction by convention is normalized to length 1. So it makes sense to assume that vectors have length 1, but that is not necessary for the definition to work. Oleg Alexandrov (talk) 21:13, 3 May 2006 (UTC)

I quite agree that it's not necessary for the definition to work, however the standard implementation and usage up till differential equations emphatically states normalized. So, we should at least say so in the definition on the page, as you did. How it stood, it could be mis-read and if someone noted JUST the formula ... they would be wrong on calculus exams. -- Dbroadwell 22:46, 3 May 2006 (UTC)
• I've got a question if a function is differentiable for any vector included on X-axis (let,s say v=(1,0)) and differentiable for any vector on Y-axis then the function is differentiable on any direction of the plane (X,Y) i think this is similar to the way in which 'Cauchy-Riemann' equations are obtained.
• I think that is important that the norm of the vector must be unitary:

If one takes two parallels non-unitary vectors, $\vec v$ $\vec w$, such that $||\vec v||\neq ||\vec w||$, then one has

$D_{\vec{v}}{f}(\vec{x}) = \nabla f(\vec{x}) \cdot \vec{v} = ||\nabla f(\vec{x})|| \; ||\vec{v}|| \cos(\theta)$

and

$D_{\vec{w}}{f}(\vec{x}) = \nabla f(\vec{x}) \cdot \vec{w} = ||\nabla f(\vec{x})|| \; ||\vec{w}|| \cos(\theta)$

where $\theta$ is the angle between the gradient and the vectors (remember they are parallels). The directional derivative depends on direction, then one must verify

$D_{\vec{v}}{f}(\vec{x}) = D_{\vec{w}}{f}(\vec{x})$

that is

$||\nabla f(\vec{x})|| \; ||\vec{v}|| \cos(\theta) = ||\nabla f(\vec{x})|| \; ||\vec{w}|| \cos(\theta)$
$||\vec{v}|| = ||\vec{w}||$

Absurd, it was supposed that $||\vec v||\neq ||\vec w||$.

In addition, it seems obvious that the directional derivative cant depend on the vector that one chooses. Because of that I think that is not by convention to choose an unitary vector, if you don´t, your result is a function of the norm of the vector chosen. Sorry about my english, Im not used to express myself in this language.
--anon

I see where Oleg is coming from, differentiating with respect to 2x is different to differentiating with respect to x, however I don't believe that "derivative along a vector" is an accurate definition of the directional derivative. I believe that it is "derivative in the *direction* of a vector" ... why else would it be called the DIRECTIONal derivative? I would like to follow what every single text book I have ever read says and normalize v.

Oleg, please look at Dbroadwell's links above and if you'd like, find a reference to what you call the directional derivative.129.78.64.101 02:23, 4 October 2007 (UTC)

I'm pretty sure the directional derivative, when defined in the direction of a nonunit vector, is supposed to be the same as the directional derivative in the direction of the unit vector which is parallel to the given vector. This agrees with wolfram mathworld, so I'm editing this page to that effect. I'm also putting the unit vector definition first, as that everyone agrees on. If some authors define the directional derivative to be what the article had before I edited it, could we include that fact with citation, but as an alternate definition rather than as the only one? 75.22.201.232 (talk) 12:55, 24 March 2010 (UTC)

$D_{\vec{v}}{f}(\vec{x}) = \nabla f(\vec{x}) \cdot \vec{v} = \nabla_v f(\vec{x})$

Is the second equal sign true? I assume, but I didn't see it explicitly mentioned in the article. User:Nillerdk (talk) 17:01, 29 January 2008 (UTC)

Thanks for pointing out the inconsistency in notation. The issue has been addressed. Silly rabbit (talk) 17:07, 29 January 2008 (UTC)

--> Just a question, how would i go about determining the second order directional derivative, if the answer were to be fuu=fxx.a^2 +fyy.b^2, isn't there going to be cases that dont hold. for instance if i take a sheet and fix a point in it's middle, pull the corners upwards and pull the midpoints of each edge down wards. Would i have not created a case where fxx,fyy= -ve and fuu = +ve?

The second order derivative along u = (a,b) is given by
$D^2f\{\mathbf{u},\mathbf{u}\} = a^2 f_{xx} + 2ab f_{xy} +b^2f_{yy}$
provided f is twice continuously differentiable. The formula you gave is missing the crossterm. siℓℓy rabbit (talk) 14:27, 15 June 2008 (UTC)

in my text book they jumped to the conclusion that [let d = delta] dz= (fx+e1)dx + (fy +e2)dy, (not that its hard to believe). instead, i proved the directional derivative equation by using continuity and following a pathway along f(x,y)dx and f(x+dx,y)dy. is there a similar method for deriving the second derivative. let me know if you have a link etc. i think it would be nice to see a wikipedia page on the second derivative test one day, i have not been able to find one. —Preceding unsigned comment added by 122.110.28.250 (talk) 09:29, 22 June 2008 (UTC)

See Second partial derivative test and Second derivative test. siℓℓy rabbit (talk) 13:36, 22 June 2008 (UTC)

## differential derivative of a point

Bold text —Preceding unsigned comment added by 220.225.127.88 (talk) 07:32, 17 April 2009 (UTC)

## Image

Hello,

Here is a link to an image which may be useful.

By the way, the french definition does not utilize normalized vector, and, more important, I do not understand why the limit is calculated with h taking only positive values

Sorry for my bad english —Preceding unsigned comment added by 90.43.213.76 (talk) 21:03, 25 March 2010 (UTC)

I think the French version does use normalized vectors. The French version of the article says, "On parlera de dérivée directionnelle de f au point u dans la direction de h lorsque le vecteur h est unitaire." So it's only willing to use the name "directional derivative" when the vector is unit. Otherwise it uses the term "La dérivée de f au point u selon le vecteur h," i.e. the derivative of f at u along the vector h.
As for the restriction to positive values - I've seen both definitions, but it's arguable that the directional derivative of f in the direction h should take into account how much f changes when you move in the direction h, but not in the reverse direction. That's all the restriction to positive values does.
In any case, I think what's going on here is that there are multiple different conventions. In the case of a differentiable function and a unit vector, they all agree, but otherwise there are slight differences between conventions.
Also, your English was flawless. 75.22.201.232 (talk) 19:26, 27 March 2010 (UTC)
Thank you for this convincing answer; I had corrected the French version, after various readings. (Asram) —Preceding unsigned comment added by 90.43.214.46 (talk) 23:48, 3 April 2010 (UTC)

## normalisation

The definition

$\nabla_{\vec{v}}{f}(\vec{x}) = \nabla f(\vec{x}) \cdot \frac{\vec{v}}{|\vec{v}|}$

appearing at the beginning of the article is in direct contradiction with the definition given in the "differential geometry" section. Tkuvho (talk) 10:01, 3 May 2010 (UTC)

## Normal derivative

I can't make sense of

$\frac{\partial f}{\partial \mathbf{v}}\cdot\mathbf{u} = Df(\mathbf{v})$

I suppose there has something gone wrong --Trigamma (talk) 21:57, 28 May 2010 (UTC)

The partial derivative-like notation on the left means the gradient of the scalar-valued function f, which happens to be the total derivative of a scalar-valued function with respect to all its variables. If f were vector-valued, it would be instead the Jacobian, which is the total derivative of one vector with respect to the other. The other notations in that equation make it clear that this is what is meant.--Jasper Deng (talk) 08:33, 23 November 2013 (UTC)

## slight clean up

Minor changes:

1. add a couple more sources,
2. introduce better numbering scheme (indented and non-bracketed),
3. made the main properties stand out more,
4. made vector notation consistent all the way through (it was a mix of bold + overarrow, typographically bold is slightly easier)

13:23, 13 April 2012 (UTC)

## Verifiability of definition

As may be seen from the threads above, there seems to be a lot of confusion about the definition of the directional derivative. It may be noted that the citations given are very weak.

The only sensible definition (i.e. one that is mathematically useful and elegant) does not involve any reference to normalization. My impression is that the most notable definition (i.e. generally used definition in authoritative texts) is this one, though I do not have many of these texts to hand. There are many reasons for this claim, to the extent that I would argue for the relegation of the use of the normalized vector to a footnote. A major argument for this is that in the context of functions over a set of coordinates and related contexts, the norm of a vector is usually not defined but the directional derivative remains an extremely useful concept. Even where an (indefinite) metric tensor is defined as in the physics of general relativity, the definition runs into serious difficulties with no apparent benefits.

My guess is that the weaker definition arose as the tail wagging the dog: a term was needed for the very useful construct that generalizes the partial derivative. An obvious term would be the "directional derivative", even if this may be a partial misnomer, which then was interpreted literally.

I would appreciate authoritative texts being cited by those who have access to them. — Quondum 07:15, 30 October 2012 (UTC)

I have an (elementary) textbook defining it using unit vectors, but it only considers steady vector fields and does not go beyond ordinary 3-dimensional space.--Jasper Deng (talk) 08:34, 23 November 2013 (UTC)