Differential geometry question

[Pacopag]

Homework Statement

Can some one please explain to me how to show that
J^{\alpha}{ }_{;\alpha}={1\over{\sqrt{-g}}}\partial_\alpha(\sqrt{-g}J^\alpha)

Homework Equations

\Gamma^\gamma{}_{\alpha\beta}={1\over 2}g^{\gamma\delta}(g_{\delta\alpha,\beta}+g_{\delta\beta,\alpha}-g_{\alpha\beta,\delta})
and
\partial_\alpha\sqrt{-g}=-{1\over 2}\sqrt{-g}g^{\mu\nu}g_{\mu\nu,\alpha} (I think this is correct).

The Attempt at a Solution

Here's what I've tried
J^{\alpha}{ }_{;\alpha}=g^{\alpha\beta}J_{\beta;\alpha}
=g^{\alpha\beta}(J_{\beta,\alpha}-J_\gamma\Gamma^\gamma{}_{\alpha\beta})
=g^{\alpha\beta}(J_{\beta,\alpha}-{1\over 2}J_\gamma g^{\gamma\delta}(g_{\delta\alpha,\beta}+g_{\delta\beta,\alpha}-g_{\alpha\beta,\gamma})).
Now turning to the other side
{1\over{\sqrt{-g}}}\partial_\alpha(\sqrt{-g}J^{\alpha})
={1\over{\sqrt{-g}}}(J^\alpha\partial_\alpha\sqrt{-g}+\sqrt{-g}\partial_\alpha J^\alpha)
={1\over{\sqrt{-g}}}(-{1\over 2}\sqrt{-g}g^{\mu\nu}g_{\mu\nu,\alpha}J^\alpha+\sqrt{-g}J^\alpha{}_{,\alpha})
Then cancel the sqrt(-g). But here I'm stuck.

 

[Pacopag]

Never mind. I got it. Just had to turn the crank a little. Cheers.