Prove that two transformation laws of the Christoffel symbols are the same

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Homework Statement



Prove that the transformation law

\Gamma^{\sigma '}_{\lambda '\rho '}=\frac{\partial x^\nu}{\partial x^{\lambda '}}\frac{\partial x^\rho}{\partial x^{\rho '}}\frac{\partial x^{\sigma '}}{\partial x^{\mu}}\Gamma^{\mu}_{\nu\rho}+\frac{\partial x^{\sigma '}}{\partial x^{\mu}}\frac{\partial^2 x^\mu}{\partial x^{\lambda '}\partial x^{\rho '}}

is equivalent to

\Gamma^{\sigma '}_{\lambda '\rho '}=\frac{\partial x^\lambda}{\partial x^{\lambda '}}\frac{\partial x^\rho}{\partial x^{\rho '}}\frac{\partial x^{\sigma '}}{\partial x^{\sigma}}\Gamma^{\sigma}_{\lambda\rho}-\frac{ \partial x^\mu}{\partial x^{\lambda '}}\frac{\partial x^{\lambda}}{\partial x^{\rho '}}\frac{\partial^2 x^{\sigma '}}{\partial x^{\mu}\partial x^{\lambda}}

The Attempt at a Solution



The first term is easy, just relabel the dummy indices \nu \rightarrow \lambda and \mu \rightarrow \sigma. But for the rest of the problem, I have no clue what to do.
 
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Maybe this is too late. It's a trick. Differentiate the far left and right of

\delta^{\lambda '}_{\mu '} = \frac{\partial x^{\lambda '}}{\partial x^{\mu '}} = \frac{\partial x^{\lambda '}}{\partial x^\rho} \frac{\partial x^\rho}{\partial x^{\mu '}}

with respect to x^{\alpha '}.
 
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