Derive Gauss formula $$x_{ij}=\Gamma^k_{ij}x_k+L_{ij}\textbf{n}\quad\textrm{(in Einstein notation)}$$ The coefficients $\Gamma^k_{ij}$ where $i, j, k = 1, 2$ are called Christoffel symbols and the coefficients $L_{ij},i,j = 1, 2$ are called the coefficients of the second fundamental form.
I know one way to get those equations like, letting $x_u,x_v$ and $\textbf{n}$ are linearly independent and may write
$$ \begin{align} x_{uu}&=a_1x_u+a_2x_v+a_3\textbf{n}\\ x_{uv}&=b_1x_u+b_2x_v+b_3\textbf{n}\\ x_{vv}&=c_1x_u+c_2x_v+c_3\textbf{n} \end{align} $$ But the problem is, in this way to get the coefficients are very tedious and computational heavy. Is there any other way to get those equations with less work?
Thanks in advance.
Update
proof : The three vectors $\mathbf{x}_{\mathrm{u}}, \mathbf{x}_{\mathrm{v}}$ and $\mathrm{N}$ are linearly jependent. so they span the three dimensional vector space. $$ \begin{aligned} x_{uu}&=a_1x_u+a_2x_v+a_3\textbf{N}\\ x_{uv}&=b_1x_u+b_2x_v+b_3\textbf{N}\\ x_{vv}&=c_1x_u+c_2x_v+c_3\textbf{N}\\ &\therefore x_{uu} \cdot N=a_3 \Rightarrow e=a_3 \cdots \cdots(4) \\ &x_{uv} \cdot N=b_3 \Rightarrow f=b_3 \cdots \cdots(5) \\ &x_{vv} \cdot N=c_3 \Rightarrow g=c_3 \cdots \cdots(6) \end{aligned} $$ We know $E=x_u \cdot x_u, \quad F=x_u \cdot x_v, \quad G=x_v \cdot x_v$ $$ \mathrm{e}=\mathrm{x}_{\mathrm{uu}} \cdot \mathbf{N}, f=\mathrm{x}_{\mathrm{uv}} \cdot \mathbf{N}, \mathrm{g}=\mathrm{x}_{\mathrm{vv}} \cdot \mathbf{N} $$ Differentiating $E=x_u \cdot x_u$ w.r.t. $u$, we have, $$ \begin{aligned} & E_{u}=x_{uu} \cdot x_{u}+x_{u} \cdot x_{uu}=2 x_{uu} \cdot x_u \\ \Rightarrow & x_{uu} \cdot x_{u}=\frac{1}{2} E_u \cdots \cdots(7) \end{aligned} $$ Again, differentiating $\mathrm{E}=x_{u} \cdot x_u$ w.r.t. v, we get $$ \begin{aligned} & E_v=x_{uv} \cdot x_{u}+x_{u} \cdot x_{uv}=2 x_{uv} \cdot x_{u} \\ \Rightarrow & x_{uv} \cdot x_{u}=\frac{1}{2} E_v \cdots \cdots(8) \end{aligned} $$ Also, differentiating $\mathrm{G}=\mathbf{x}_{\mathrm{v}} \cdot \mathbf{x}_v$ w.r.t. $u$ and $\mathrm{v}$, we respectively $$ \begin{aligned} & G_u=x_{uv} \cdot x_v+x_v \cdot x_{uv} \text { and } G_v=x_{vv} \cdot x_v+x_v \cdot x_{vv} \\ \Rightarrow & G_u=2 x_{uv} \cdot x_v \\ \Rightarrow & x_{uv} \cdot x_v=\frac{1}{2} G_u \cdots \cdots \cdot(9) \quad \Rightarrow x_{vv} \cdot x_v=\frac{1}{2} G_v \cdots \ldots \cdot(10) \end{aligned} $$
Update 2
So, there's no way to avoid this type of calculation without introducing some advance stuff. I read the proof once again and notice I couldn't manage to get ideas for the following lines:
$x_{u u}=\frac{G E_u-2 F F_u+F E_v}{2\left(E G-F^2\right)} x_u+\frac{2 E F_u-E E_v-F E_u}{2\left(E G-F^2\right)} x_v+e N$
$x_{u v}=\frac{G E_v-F G_u}{2\left(E G-F^2\right)} x_u+\frac{E_u-F E_v}{2\left(E G-F^2\right)} x_v+f N$
$\mathbf{x}_{\mathrm{vv}}=\frac{2 \mathrm{GF}_{\mathrm{v}}-\mathrm{GG}_u-\mathrm{FG}_{\mathrm{v}}}{2\left(\mathrm{EG}-\mathrm{F}^2\right)} \mathbf{x}_{\mathrm{u}}+\frac{\mathrm{EG}_{\mathrm{v}}-2 \mathrm{FF}_{\mathrm{v}}+\mathrm{FG}_{u}}{2\left(E G-\mathrm{F}^2\right)}+\mathrm{g} N$
These equations can be written as $$ \begin{aligned} &x_{u u}=\Gamma_{11}^1 x_{11}+\Gamma_{11}^2 x_v+e N \\ &x_{u v}=\Gamma_{12}^1 x_{11}+\Gamma_{12}^2 x_v+f N \\ &x_{v v}=\Gamma_{22}^1 x_{11}+\Gamma_{22}^2 x_v+g N \end{aligned} $$
I didn't understand how they replace those coefficients with chistoffel symbols.