Prove that $L_Adf(X) = X(Af)$ Prove that $L_Adf(X) = X(Af)$ for any vector fields $A,X \in \mathfrak{X}(M)$ and any scalar field $f \in \mathfrak{F}(M).$
Any hint? I have no idea where to start :/
 A: Use that the Lie-derivative satisfies the product rule $L_{A}(B \otimes C)=(L_{A}B) \otimes C+ B \otimes (L_{A}C)$ and the fact that it commutes with the contraction $\underbrace{}$ to compute both sides of $$L_{A}(df(X)) = L_{A}(\underbrace{df \otimes X})$$
in a different manner and conclude what $(L_{A}df)(X)$ is.
A: This can be seen as an application of the Cartan's magic formula (see e.g. here)
$$
L_A \omega = i_A \mathrm{d} \omega + \mathrm{d} (i_A \omega)
$$
Indeed, for the 1-form $\omega = \mathrm{d}f$ we have
$$
L_A  \mathrm{d}f (X) = \mathrm{d} \mathrm{d}f (A,X) + \mathrm{d} (\mathrm{d}f (A)) (X) = X(A f)
$$
using $\mathrm{d}^2 = 0$ and $Xf = \mathrm{d}f(X)$.
A: After some work I've come up with the following solution:
$L_A(df(X))= L_A(C_1^1(df \otimes X)) = C^1_1(L_A(df \otimes X)) = 
C^1_1(L_Adf \otimes X + df \otimes L_A X))= L_Adf (X) + df ([A,X])$
Then we obtain that $L_Adf(X)= L_A(df(X))- df([A,X]).$ 
$L_Adf(X) = A(Xf)-(A(Xf)-X(Af)) = X(Af).$
Here I use that Lie derivative of tensor field commutes with contraction of the tensor,and the Liebniz Rule, for sure.
At the first look the thing I got stucked was not distinguishing the difference between $L_Adf(X)= L_A(df(X))$. I hope now it is clear for whom has the similar confusion.
