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Consider smooth map $f:\mathbb{R}^{n}\to \mathbb{R}$, let $a \in \mathbb{R}^{n}$ be any point, $X \in T_{a}\mathbb{R}^{n}=\mathbb{R}^{n}$ be tangent vector at point $a$.

I have probably proven the following:

Claim: $df_{a}(X)=\frac{d}{dt}|_{t=0}f(a+Xt)$, where $d$ is the tangent map or derivation of $f$.

""Proof"": Let $h:\mathbb{R}\to \mathbb{R}$ be some smooth function. Let's evaluate $df_{a}(X)$ at $h$.

$df_{a}(X)h = X|_{a}(hf)$ (this means directional derivative by $X$ at point $a$. also this point follows from definition)

Now by chain rule, we obtain $X|_{a}(hf)=\frac{d}{dt}|_{t=f(a)}h \cdot X|_{a}f$.

I.e. $df_{a}(X) = X|_{a}f \cdot \frac{d}{dt}|_{t=f(a)}$.

Because we identify real numbers $x \in \mathbb{R}$ with one dimensional derivations $x \cdot \frac{d}{dt}|_{t=f(a)}$, we conclude:

$df_{a}(X) = X|_{a}f$

Which by the definition equals to $\frac{d}{dt}|_{t=0}f(a+Xt)$. Q.E.D.


The whole reasoning seem clumsy and far fetched to me. If not wrong. Is my proof correct? What would you suggest to change/improve? Thanks in advance.

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You proof is correct. However, there are various equivalent definitions of the tangent space $T_a\mathbb{R}^{n}$. It is isomorphic to $\mathbb{R}^{n}$, but you should make precise which definition you use and how the isomorphism $T_a\mathbb{R}^{n} \to \mathbb{R}^{n}$ is given.

Obviously you use the description via derivations. In this case $X \in \mathbb{R}^{n}$ is identfied with the directional derivation $D_{a,X}$ at $a$ in direction $X$ which you write as $X \mid_a$. In the one-dimensional case you say correctly that $x$ is identified with $x \cdot \frac{d}{dt} \mid_{f(a}$ which is the directional derivation $D_{f(a),x} = x \mid_{f(a)}$. Your claim should then precisely be understood as follows:

Under the above identications of tangent spaces, $df_{a}(X)=\frac{d}{dt}|_{t=0}f(a+Xt) = D_{a,X}(f) = X \mid_a(f)$.

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