Laplacian in $\mathbb{R}^n$ expressed with Christoffel symbols I thought I'd try my luck with another question, so here it goes:
I have to show that if $(y^1,...,y^n):\mathbb R^n \rightarrow \mathbb R^n$ is a diffeomorphism and $f\in C^\infty(\mathbb R^n)$. Then the Laplacian in $\mathbb R^n$ $\nabla_{\mathbb R^n} = \sum_{j=1}^n(\frac{\partial}{\partial x^j})^2$  on the composite function $f$ is
$\nabla_{\mathbb R^n} f(y^1(x^1,...,x^n),...y^n(x^1,...x^n)) = (\widetilde{\nabla}f)(y^1(x^1,...,x^n),...,y^n(x^1,...,x^n))$
where $\widetilde{\nabla} = \sum_{i,j=1}^n g^{ij}((\frac{\partial}{\partial y^i} \frac{\partial}{\partial y^j}) - \sum_{k=1}^n \Gamma_{ij}^k \frac{\partial}{\partial y^k})$
and $g^{ij}=\sum_{l=1}^n \frac{\partial y^i}{\partial x^l}\frac{\partial y^j}{\partial x^l}$
and  $\Gamma_{ij}^k = \sum_{l=1}^n \frac{\partial y^k}{\partial x^l} \frac{\partial ^2 x^l}{\partial y^i \partial y^j}$ are the Christoffel symbols
So far I've tried to just write out the Laplacian but I got stuck on the second derivative.
My Calculations:
$\nabla_{\mathbb R^n} f(y^1(x^1,...,x^n),...y^n(x^1,...x^n)) = \sum_{j=1}^n (\frac{\partial}{\partial x^j})^2(f) = \sum_{j=1}^n \frac{\partial}{\partial x^j}(\frac{\partial f}{\partial y^j}\frac{\partial y^j}{\partial x^j}) = \sum_{j=1}^n \frac{\partial ^2 f}{\partial (y^j)^2}(\frac{\partial y^j}{\partial x^j})^2 + \frac{\partial f}{\partial y^j}\frac{\partial ^x y^j}{\partial (x^j)^2}$ 
but then I'm not quite sure what I should do next. Can anyone help me with this? Thanks in advance!
 A: This is a computation in which you can largely follow your nose. 
A direct evaluation by chain and product rules gives
$$ \nabla_{\mathbb{R}^n}f(y^1(\ldots), \ldots , y^n(\ldots)) = \sum_{i,j ,k = 1}^{n} \frac{\partial^2 f}{\partial y^i \partial y^j} \frac{\partial y^j}{\partial x^k} \frac{\partial y^i}{\partial x^k} + \sum_{i,j = 1}^n \frac{\partial f}{\partial y^j} \frac{\partial^2 y^j}{\partial x^i \partial x^i} $$
The first term in the expression is clearly equal to $ \sum_{i,j} g^{ij} \frac{\partial^2 f}{\partial y^i \partial y^j}$ as desired. To read off the second term we use the change of variables formula for Christoffel symbols. Using that the Christoffel symbol for the Euclidean metric in the $x$ (standard) coordinate system vanishes, the change of variable formula (look at here but switching the $y$ and $x$ coordinates) gives 
$$ 0 = \sum_{ij} \frac{\partial y^i}{\partial x^p} \frac{\partial y^j}{\partial x^q} \Gamma^k_{ij} + \frac{\partial^2 y^k}{\partial x^p \partial x^q}$$
which when substituted into the second term gives the correct Christoffel symbol expresion. 
