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If $f$ is a function from $\mathbb{R}^n$ to $\mathbb{R}$, then its derivative at a point $\mathbf{u}$, $f'(\mathbf{u})$ is a linear transformation from $\mathbb{R}^n$ to $\mathbb{R}$. But we know that any linear transformation $T$ from $\mathbb{R}^n$ to $\mathbb{R}$ is of the form $T\mathbf{x}=\mathbf{x}\cdot \mathbf{y}$. Hence $f'(\mathbf{u})\mathbf{x}=\mathbf{x}\cdot \mathbf{y}$ for some $\mathbf{y}\in \mathbb{R}^n$. Would there be any connection between the vectors $\mathbf{u}$ and $\mathbf{y}$?

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There is absolutely no relation between $u$ and $y$. However $f'(u)=T$ and $y$ correspond to each other through the implicit euclidean structure of $\mathbb R^n$.This is reflected in the terminology covariant vs. contravariant vectors un Differential Geometry or bra and ket vectors in Quantum Mechanics. –  Georges Elencwajg Jan 27 '12 at 9:44
    
@GeorgesElencwajg I don't understand what you mean by "However $f'(u)=T$ and $y$ correspond to each other through the implicit euclidean structure of $\mathbb{R}^n$". Could you please elaborate on that? –  Ashok Jan 27 '12 at 10:10
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The euclidean structure on $\mathbb R^n$ yields a linear isomorphism $euc: \mathbb R^n \to L(\mathbb R^n,\mathbb R): y\mapsto euc_y$ where the linear form $euc_y \in L(\mathbb R^n,\mathbb R)$ is defined by $euc_y(x)=x\cdot y$. In your case, the vector $y=grad f(u)=\left (\frac{\partial f}{\partial x_1}(\mathbf{u}),\frac{\partial f}{\partial x_2}(\mathbf{u}),\frac{\partial f}{\partial x_3}(\mathbf{u}),\cdots \right)\in \mathbb R^n$ corresponds to the linear form $euc_y=euc _{grad f(u)}= f'(u) \in L(\mathbb R^n,\mathbb R)$. (cf. NumberFour's absolutely correct answer) –  Georges Elencwajg Jan 27 '12 at 10:35
    
Thanks @GeorgesElencwajg. This is very informative. –  Ashok Jan 27 '12 at 10:43
    
My point was to emphasize the difference of culture with physicists, who sometimes use a terminology alluding to different kinds of vectors (bra's and ket's, for example) where we emphasize the contrast between vectors and linear forms. Also, for psychological reasons, gradients seem more concrete than derivatives. All this is quite shallow; one should just know that these small nuances exist and understand them. –  Georges Elencwajg Jan 27 '12 at 10:44
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up vote 2 down vote accepted

I dont know if this answers your question, but in case of transformation from $\mathbb{R}^n $ to $\mathbb{R}$ the f'($\mathbf{u}$) is the gradient of f at the point $\mathbf{u}$:

$$ \left (\frac{\partial f}{\partial x_1}(\mathbf{u}),\frac{\partial f}{\partial x_2}(\mathbf{u}),\frac{\partial f}{\partial x_3}(\mathbf{u}),\cdots \right)$$ which is a 1xN matrix - or your vector $\mathbf{y}$. This is the result of the famous Riesz representation theorem. So applying some vector $\mathbf{x}$ to this matrix yields a directional derivative in direction $\mathbf{x}$ at the point $\mathbf{u}$. Mathematically written: $$ f'(\mathbf{u})\mathbf{x} = \nabla f(\mathbf{u})\cdot\mathbf{x} $$

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Thanks!I just realized that. Accordingly, I have deleted my comment. –  Ravi Donepudi Jan 27 '12 at 9:51
    
I know this. I was wondering whether $\mathbf{u}$ should be equal to $\mathbf{y}$ or something like that. Any way, thanks. I will wait some more time to see whether there is some affirmative answer before accepting yours. –  Ashok Jan 27 '12 at 10:07
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