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From: fdm on 5 Sep 2009 15:14
As I understand the gradient of 'f' is a vector pointing in the direction
with the most growth while the derivative of 'f' is "how much" the growth is
actually increased in that direction for each unit of increase in the
domain. Is that correct?



"Ray Vickson" <RGVickson(a)shaw.ca> wrote in message
news:5217d1fe-9bd6-46b0-a0cb-dd9bfb3fd025(a)2g2000prl.googlegroups.com...
On Sep 5, 10:50 am, "fdm" <f...(a)gk.com> wrote:
> Would it be correct to say that the i'th row corresponds to the gradient
> of
> the i'th functin?

I don't know who you are addressing, but the answer is YES. That is
obvious, just from the definition.

R.G. Vickson

>
> "Gordon Stangler" <gordon.stang...(a)gmail.com> wrote in message
>
> news:a3acf158-425e-428a-89dd-8342e7981166(a)s31g2000yqs.googlegroups.com...
> On Sep 5, 10:14 am, "fdm" <f...(a)gk.com> wrote:
>
>
>
> > The two variable function:
>
> > f(x,y)
>
> > has the gradient (vector):
>
> > F(x,y) = (\par x, \par y)
>
> > I have read that the jacobian is the generalization of the gradient and
> > that
> > it always has the form of a matrix. I have also read that the jacobian
> > is
> > the first order partial derivative of a multivariable function but that
> > is
> > also the definition of the gradient.
>
> > So what is the difference between the gradient and the jacobian?
>
> > And what is the jacobian of f above?
>
> The Jacobian is explained quite nicely
> here:http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant
> There are even examples if you need them.

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