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For $f \colon \mathbb{R}^n \to \mathbb{R}$ the gradient of $f$ is a vector field over $\mathbb{R}^n$ defined as $$\nabla f := \sum_{i=1}^n \frac{\partial f}{\partial x_i}\frac{\partial}{\partial x_i}.$$

How I can prove that $df_p(v) = \langle v, \nabla_p f \rangle$?

If $h \colon \mathbb{R}^n \to \mathbb{R}$ is another differential map, how to compute $[\nabla f, \nabla g]h?$

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  • $\begingroup$ What do you mean by your last question? $\endgroup$
    – Gibbs
    Nov 5 '18 at 15:26
  • $\begingroup$ Lie brackets of vector fields! :) $\endgroup$
    – LH8
    Nov 5 '18 at 15:28
  • $\begingroup$ So do you want to find $[\nabla f, \nabla g]h$ explicitly? $\endgroup$
    – Gibbs
    Nov 5 '18 at 15:28
  • $\begingroup$ Yes, that´s right $\endgroup$
    – LH8
    Nov 5 '18 at 15:33
  • $\begingroup$ See math.stackexchange.com/questions/370851/… for computation of Lie bracket of vector fields. $\endgroup$
    – edm
    Nov 5 '18 at 15:40
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Since $$v = \sum_{k=1}^n v^k\frac{\partial}{\partial x^k}$$ then by definition of differential of a function you get \begin{align} df_p(v) & = \sum_{i=1}^n \frac{\partial f}{\partial x^i}_{p}dx^i(v) = \sum_{i=1}^n \frac{\partial f}{\partial x^i}_{p} \sum_{k=1}^n v^k dx^i\left(\frac{\partial}{\partial x^k} \right) \\ & = \sum_{i=1}^n \frac{\partial f}{\partial x^i}_{p} \sum_{k=1}^n v^k \delta^i_k = \sum_{i=1}^n \frac{\partial f}{\partial x^i}_{p} v^i = \langle v, \nabla_p f \rangle, \end{align} where $\langle \cdot, \cdot \rangle$ is the Euclidean metric on $\mathbb{R}^n$ and $\delta^i_k$ the Kronecker delta.

A hint for the second part: if $X = \sum_i X^i \partial_i$ and $Y = \sum_k Y^k \partial_k$ are two vector fields, then if $h$ is a smooth function \begin{align} [X,Y]h& = X(Yh)-Y(Xh) \end{align} with $Xh = \sum_i X^i \partial_i h$, and similarly for $Yh$.

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