Ytukyg

In Euclidean space $ mathbb{R}^n $, let $ X_j=frac{partial}{partial x_j} $, and for every $ X, Yinmathfrak{X}(mathbb{R}^n) $, if
$$ X=sum_{i}f_iX_i=(f_1, f_2, cdots, f_n),quad Y=sum_{j}g_jX_j=(g_1, g_2, cdots, g_n), $$
then
begin{align*}
nabla_XY&=sum_jX(g_j)X_j+sum_{i, j}f_inabla_{X_i}X_j=sum_jX(g_j)X_j\
&=(X(g_1), X(g_2),cdots, X(g_n))\
&=operatorname{d} Y(X).
end{align*}

This is from my textbook. I don't understand why $ (X(g_1), X(g_2),cdots, X(g_n))=operatorname{d}Y(X) $? ($ operatorname{d} $ is the tangent map).

In $M=mathbb{R}^n$ we have the global chart given by the identity map $phi:xin M mapsto x in mathbb{R}^n$. For every point $p$ in $M$ we have the basis of the tangent vectors given by this chart $frac{partial}{partial varphi_i}|_p$ with $1leq i leq n$. They are the usuar directional derivatives, $frac{partial}{partial varphi_i}|_p=frac{partial}{partial x_i}|_p$. So for every smooth function $f:Mto mathbb{R}$ and for every $i$, we have $frac{partial}{partial varphi_i}|_p(f)=frac{partial f}{partial x_i}(p)$. In your notation, $X_i=frac{partial}{partial varphi_i}|_p$.

Every smooth vector field $X$ in $mathfrak{X}(M)$ can be writen univocally as $X=sum_i^ng_ifrac{partial}{partial varphi_i}$ where $g_i:Mto mathbb{R}$ is smooth and is given by $g_i(p)=X_p(varphi_i)$ for every $i$. So we have an isomorphism
$$
X in mathfrak{X}(M)mapsto (X(varphi_1),ldots,X(varphi_n)) in mathcal{C}^infty(M,mathbb{R}^n)
$$
with inverse
$$
(g_1,dots,g_n) in mathcal{C}^infty(M,mathbb{R}^n) mapsto sum_i^ng_ifrac{partial}{partial varphi_i} in mathfrak{X}(M)
$$

Under the identifications above, consider a smooth vector field $Y=(g_1,ldots,g_n)$. It is just a smooth function $mathbb{R}^ntomathbb{R}^n$ so you can compute its differential map (or tangent map). For every $p$ in $M$, $d_pY: T_pM to T_{Y(p)}mathbb{R}^n$ is given by $d_pY(X_p)(f)=X_p(fcirc Y)$ for every $f$ in $mathcal{C}^infty(M)$. We can again use the identifications $T_pMcong mathbb{R}^n$ and $T_{Y(p)}congmathbb{R}^n$. Under this identifications $d_pY$ is just the linear endomorphism of $mathbb{R}^n$ with the differential matrix of $Y$ at $p$ as asociated matrix. That explain $(X(g_1),ldots,X(g_n))=(dg_1(X),ldots,dg_n(X))=dY(X)$.