How can I make two separate equations for Christoffel symbols give the same answer? I have been studying the covariant derivative and I'm confused by the calculation of the Christoffel symbols $\Gamma$. The equation for computing $\Gamma$ is given as:
$${\Gamma^c}_{ab} = \frac12 g^{ck} (\partial_a g_{bk} + \partial_b g_{ak} - \partial_k g_{ab} ) $$
This will produce a $\Gamma$ which is symmetric in a and b. And it transforms as:
$$\Gamma' = T \Gamma S S + \partial S T$$
Where $S$ is the transformation matrix and $T$ is the inverse. Using this equation, if we are transforming from a flat space where $\Gamma = 0$, then:
$$\Gamma' = \partial S T$$
But this equation will not always produce a $\Gamma$ which is symmetric. and so won't give the same answer as the first equation. I've tried this with the following transformation matrix in an $(x,y)$ coordinate system:
$$ S = \begin{bmatrix}1  & 0 \\ 0 & x \end{bmatrix} $$
This produces a metric tensor of:
$$ g' = gSS = \begin{bmatrix}1  & 0 \\ 0 & x^2 \end{bmatrix}$$
This is the same metric tensor as for polar coordinates (with $x$ instead of $r$). Using the first equation, the Christoffel symbols end up as:
$$ \Gamma' = \begin{bmatrix}0 & 0 \\ 0 & -x\end{bmatrix}
   _{\begin{bmatrix}0 & \frac1x \\ \frac1x & 0\end{bmatrix}}$$
But using the second equation I get:
$$ \Gamma' = \partial S T = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}
   _{\begin{bmatrix}0 & 0 \\ \frac1x & 0\end{bmatrix}}$$
Why are they different? I must be missing something, perhaps obvious, but I've researched it and calculated it many times and can't find the problem. 
 A: Thanks to MBN for the comment - which is also pretty much the answer. Which is basically that I can not make the two equations give the same result, because the  $S$ which I was using is invalid.
I was thinking of $S$ in terms of a matrix of coordinate changes in a vector tangent space. The transformation I was trying to achieve was:
$$x' = x, y' = xy $$
So then $S$ could be used to transform from $(x, y)$ coordinates to $(x',y')$:
$$ \begin{bmatrix}x' \\ y' \end{bmatrix} = S \begin{bmatrix}x \\ y \end{bmatrix} =  \begin{bmatrix}1 & 0 \\ 0 & x \end{bmatrix} \begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & xy \end{bmatrix} $$
But the $S$ used above in the formula for $\Gamma$ is the Jacobian matrix of partial derivatives:
$$S = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \\ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ y & x \end{bmatrix} $$
Using that $S$ ends up with a completely different $g$ and $\Gamma$. So my confusion was in not understanding the difference between coordinate transformation matrices and Jacobians, and therefore using an impossible $S$.
