Christoffel Symbol in terms of $\sqrt{|g|}$ I'm trying to understand the conversion of 
$$ \Gamma^{\mu}_{\mu \nu} = \frac{1}{\sqrt{|g|}} \partial_{\nu}(\sqrt{|g|}) $$
where $g=\det{g_{uv}}$ 
Working it out, I get to this form of the connection
$$ \Gamma^{\mu}_{\mu \nu} = \frac{1}{2} g^{\mu\lambda} \partial_{\nu} g_{\mu\lambda}$$
from here, I'm stuck on how to apply the determinant to the metric and remove the $1/2$
From my general relativity text book, Carroll takes the coordinate transformation and applies the determinant to both sides.
$$ g_{\mu' \nu'} = \frac{\partial x^{\mu}}{\partial x^{\mu'}} \frac{\partial x^{\nu}}{\partial x^{\nu'}} g_{\mu \nu}   $$
$$ g(x^{\mu'}) =  \Big| \frac{\partial x^{\mu'}}{\partial x^{\mu}} \Big|^{-2}\ g(x^{\mu})   $$
Why is the Jacobian raised to the power of -2? And where does the square root come in?
 A: To me that seems to be two completely different questions. First of all, 
$$\tag{1} \Gamma^{\mu}_{\mu \nu} = \frac{1}{\sqrt{\det g}} \partial_{\nu}(\sqrt{\det g})$$
is equivalent to 
$$\Gamma^{\mu}_{\mu \nu} = \frac{1}{2} g^{\mu\lambda} \partial_{\nu} g_{\mu\lambda},$$
since we have the Jacobi's formula:
$$\frac{d}{dt} \log\det(A(t)) = \text{tr}(A^{-1} A'(t)).$$
This implies (note that I did not use $g$ to represent the determinant, as you do)
$$ \partial _v \det{g} = (\det g) \text{tr} (g^{-1} \partial _v g) = (\det g)g^{\mu\lambda}\partial_\nu g_{\mu\lambda}$$
and $(1)$ follows from chain rule (of course that also explain the $\frac 12$, which comes from the derivatives of $\sqrt{\cdot}$ ). 
For the second question, note that as matrix equation, 
$$g_{\mu' \nu'} = \frac{\partial x^{\mu}}{\partial x^{\mu'}} \frac{\partial x^{\nu}}{\partial x^{\nu'}} g_{\mu \nu}$$
is written as 
$$g' = J g J^T, $$
where $J$ is matrix given by $J_{\mu'\mu} = \frac{\partial x^{\mu}}{\partial x^{\mu'}}$. This implies 
$$ \det g' = \det J \det g \det J^T = \det J^2 \det g = (\det J^{-1})^{-2} \det g.$$
Since $J^{-1}$ is the matrix with coefficient $\frac{\partial x^{\mu'}}{\partial x^{\mu}}$, the last equality is justified. 
