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We define derivatives of functions as linear transformations of $R^n \to R^m$. Now talking about the derivative of such linear transformation , as we know if $x \in R^n$ , then

$A(x+h)-A(x)=A(h)$, because of linearity of $A$, which implies that $A'(x)=A$ where , $A'$ is derivative of $A$ . What does this mean? I am not getting the point I think.

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What does "what does this essentially mean" mean? – Qiaochu Yuan Jun 4 '12 at 17:15
@QiaochuYuan : It means how do i understand it. I didn't understand it . – Theorem Jun 4 '12 at 17:17
I agree with Qiaochu's objection. Anyway, I'd say: the derivative of a mapping between normed spaces is defined as "the best linear approximation" (in a sense to be quantified precisely). Clearly the best linear approximation of a linear map is the map itself. – Giuseppe Negro Jun 4 '12 at 17:18
Perhaps considering the one-dimensional case will give some clarity. Take $f(x) = ax$, which is a linear transformation from $\mathbb R$ to $\mathbb R$. Then $f'(x) = a$. – Rahul Jun 4 '12 at 18:12

$A'$, where $A$ is seen as a linear /map/, has a derivative $A$, where $A$ is now seen as a (constant) matrix..

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