Derivate of $f(x)=x$ with $x \in R^n$
how $f(x+y)=x+y=f(x)+f(y)$ and let $a \in R$ then $f(ax)=ax=af(x)$ then $f$ is a linear transformation and then $f'(x)=f(x)$ because the derivate of a linear transformation is the same funtion is this right?
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The claim that $f'=f$ for linear $f$ is false. Note that the real function $x\mapsto x$ is linear, but its derivative is $x\mapsto 1$. – Alann Rosas Dec 06 '22 at 06:16
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@AlannRosas for for functions $f:R^n \righatarrow R^m$ is not true? – Gustavo Ortega Dec 06 '22 at 06:18
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There is some misunderstanding on what the statement "derivative of a linear transformation is the same function" means. If $y=f(x)=x$ then, indeed, $dy=dx$, but the number we also call the derivative is $f'(x)=\frac{dy}{dx}=1$, not $x$. – Dec 06 '22 at 06:22
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@StinkingBishop ok i understand then the derivate is the identity matriz right? using the global derivate (jacobian matrix) – Gustavo Ortega Dec 06 '22 at 06:43
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The word derivative covers a dozen of meanings, depending on how you look at it. E.g. if we consider the Weierstraß formula for a derivative $$f(x+v)=f(x)+D_x(f)\cdot v+ r(v)$$ and assume that $f$ is linear, then we get with $r\equiv 0$ that $$f(v)=D_x\cdot v.$$ In this sense, i.e. as a function of the direction in which we differentiate, the derivative at a certain point $x$ is the original linear function $f.$ In case $f(x)=x$ this reads $$f'_{x_0}\cdot(v)=\left. \dfrac{d}{dx}\right|_{x=x_0}f(x)\cdot v = 1\cdot v=v=f(v).$$ This is meant if people say that the derivative of a linear function is the function again. It is due to the multiple uses of a derivative: the slope of the graph of $f$, linear approximation of $f$, function of the direction of differentiation (derivatives are always directional), and function of the location at which we differentiate, and it goes on if you include concepts like fiber bundles, differential forms etc.
Marius S.L.
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