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I am studying the chapter 'Several Variables' from Rudin's "Principles of Mathematical Analysis". There I found a new conception of derivative which generalizes the notion of derivatives to higher dimensions. He explained it as follows which I write here in my own languange of understanding $:$

Suppose $f:(a,b) \longrightarrow \mathbb R$ be differentiable function. Then for each $x \in (a,b)$, $f'(x)$ exists. Instead of viewing $f'(x)$ for some $x \in (a,b)$ as a real number we may treat it as a linear transformation from $\mathbb R$ into $\mathbb R$. Because for every real number $\alpha$ we can associate a linear operator $T_{\alpha}$ on $\mathbb R$ defined by $T_{\alpha} (x) = \alpha x$, $x \in \mathbb R$. Conversely, any linear operator $T$ on $\mathbb R$ can be defined as $T(x)= \alpha x$ for some $\alpha \in \mathbb R$ (we may take $\alpha=T(1)$). Hence there exist a $1-1$ correspondence from $\mathbb R$ onto $L(\mathbb R)$. Also it can be easily checked that this kind of correspondence is linear. So it's an isomorphism. Therefore we may identify $f'(x)$ by the linear operator $T_{f'(x)}$ on $\mathbb R$ defined by $T_{f'(x)} (h) = f'(x) h$, $h \in \mathbb R$. Similarly if $f : (a,b) \longrightarrow \mathbb R^m$ is a differentiable function we then find an isomorphism from $\mathbb R^m$ onto $L(\mathbb R, \mathbb R^m)$ and hence instead of identifying $f'(x)$ for some $x \in \mathbb R$ as a vector in $\mathbb R^m$ we may simply identify it as a linear transformation $T_{f'(x)}$ from $\mathbb R$ to $\mathbb R^m$ defined by $T_{f'(x)} (h) = f'(x) h$, $h \in \mathbb R$.

But for higher dimensions I fail to relate this concept. Suppose $f : E \subset_{open} \mathbb R^n \longrightarrow \mathbb R^m$ is differentiable on $E$ then how can I find the $1-1$ correspondence which is linear? I have thought about it for at least half an hour but couldn't find any satisfactory answer. Please help me at this point.

Thank you in advance.

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  • $\begingroup$ You are right. If $f : \mathbb R^n\to \mathbb R^m$ then the derivative is in $L(\mathbb R^n,\mathbb R^m)$, which is not isomorphic to $\mathbb R^n$. $\endgroup$ – amsmath Oct 14 '17 at 16:56
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If $f:\Bbb{R}^n\to\Bbb{R}^m$, then we can write $f=\langle f_1,\ldots,f_m\rangle$, where each $f_i=f_i(x_1,\ldots,x_n)$.

Write down a matrix of partial derivatives, where the $(i,j)$ entry is the derivative $\frac{\partial f_i}{\partial x_j}$. (Each row is the gradient of one of the coordinate functions.) This gives us an $m\times n$ matrix that represents a linear transformation $\Bbb{R}^n\to\Bbb{R}^m$. That transformation is the linearization of $f$ at any particular point in its domain. It corresponds to the "value" of a derivative of a $1$-variable function at a point.

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If $f : \mathbb R\to \mathbb R$ is differentiable at $x_0$, then $$ f(x_0+h) - f(x_0) = f'(x_0)h + R(h), $$ where $R(h)/h\to 0$ as $h\to 0$. This is the relation that can be raised to higher dimensions. One defines $f : \mathbb R^n\to\mathbb R^m$ to be differentiable at $x_0\in\mathbb R^n$ if there exists a linear map or matrix (both are the same) $A : \mathbb R^n\to\mathbb R^m$ such that $$ f(x_0+h) - f(x_0) = Ah + R(h), $$ where $R(h)/|h|\to 0$ as $|h|\to 0$. Note that in this case, $h$ is a vector. The matrix $A$ is then often denoted by $f'(x_0)$ or $Df(x_0)$ or $D_{x_0}f$. It always contains the partial derivatives as described in the answer of G Tony Jacobs.

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