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 $\mathbf{y}$?
Derivative of multivariable function
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analysis
multivariable-calculus
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0My 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. – 2012-01-27
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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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0I 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. – 2012-01-27