Let $M$ be a manifold and $f:M\longrightarrow \mathbb{R}$ a smooth map. Given a chart $(\mathbb{U},\phi=(x^1,x^2,\ldots,x^n))$ on $p\in M$, I define $\dfrac{\partial f}{\partial x^k}|_p:=\dfrac{\partial (f\circ \phi^{-1} )}{\partial r^k}\vert_{\phi(p)}$, where $(r^1,r^2,\ldots,r^n)$ are the standard coordinates of $\mathbb{R}^n$. I think that the value of the derivate at $p$ is indipendent from the choice of the chart $(\mathbb{U},\phi)$. How I can prove this assertion, if it is true?

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    $\begingroup$ Can you explicite this strategy? $\endgroup$ – Vincenzo Zaccaro Dec 4 '16 at 21:38
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    $\begingroup$ I don't think I quite understand what you're asking. The very definition of the operator $\partial/\partial x^k|_p$ depends on the choice of chart. If you choose a different chart $(\mathbb V, \psi = (y^1,\dots,y^n))$ whose domain contains $p$, there is no reason why the operators $\partial/\partial x^k|_p$ and $\partial/\partial y^k|_p$ should be equal. $\endgroup$ – Jack Lee Dec 4 '16 at 22:44
  • $\begingroup$ you are right, @JackLee, my bad! $\endgroup$ – L.F. Cavenaghi Dec 4 '16 at 23:06
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    $\begingroup$ @VincenzoZaccaro: Yes, exactly. $\endgroup$ – Jack Lee Dec 5 '16 at 0:16
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    $\begingroup$ @frusciante14: No, it's not enough for the charts to coincide at one point. Because the transformation law for coordinate derivatives involves the Jacobian of the transition functions, you need the first derivatives of the coordinate functions to match at $p$ also. $\endgroup$ – Jack Lee Dec 5 '16 at 0:17

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