I have been asked to show that $\nabla_{\nu} V^{\lambda} = \partial_{\nu}V^{\lambda} + \Gamma ^{\lambda}_{\mu \nu} V^{\mu}$ is a [1, 1] tensor, using the following results:
$$\tilde{\Gamma}^{\lambda}_{\mu \nu} = \frac{\partial \tilde{x}^{\lambda}}{\partial x^{\rho}} \frac{\partial^{2}x^{\rho}}{\partial \tilde{x}^{\nu} \tilde{x}^{\mu}} + \frac{\partial \tilde{x}^{\lambda}}{\partial x^{\rho}} \frac{\partial x^{\sigma}}{\partial \tilde{x}^{\nu}}\frac{\partial x^{\gamma}}{\partial \tilde {x}^{\mu}} \Gamma^{\rho}_{\gamma \sigma}$$
and
$$\tilde{\partial}_{\mu} \tilde{V}^{\nu} = \frac{\partial x^{\rho}}{\partial \tilde{x}^{\mu}} \frac{\partial \tilde{x}^{\nu}}{\partial x^{\sigma}} \frac{\partial V^{\sigma}}{\partial x^{\rho}} + \frac{\partial x^{\rho}}{\partial \tilde{x}^{\mu}} \frac{\partial^{2} \tilde{x}^{\nu}}{\partial x^{\rho} \partial x^{\sigma}} V^{\sigma}$$.
I have attempted this, but without success as I cannot seem to get the second derivative terms to cancel. I did this by trying to manipulate $\tilde{\Gamma}^{\lambda}_{\mu \nu}\tilde{V}^{\mu}$ as follows:
$(\frac{\partial \tilde{x}^{\lambda}}{\partial x^{\rho}} \frac{\partial^{2}x^{\rho}}{\partial \tilde{x}^{\nu} \tilde{x}^{\mu}} + \frac{\partial \tilde{x}^{\lambda}}{\partial x^{\rho}} \frac{\partial x^{\sigma}}{\partial \tilde{x}^{\nu}}\frac{\partial x^{\gamma}}{\partial \tilde {x}^{\mu}} \Gamma^{\rho}_{\gamma \sigma})(\frac{\partial \tilde{x}^{\mu}}{\partial x^{\alpha}}V^{\alpha}) \\= \frac{\partial \tilde{x}^{\lambda}}{\partial x^{\rho}} \frac{\partial {x}^{\sigma}}{\partial \tilde{x}^{\sigma}} \Gamma^{\rho}_{\alpha \sigma}V^{\alpha} + \frac{\partial \tilde{x}^{\lambda}}{\partial {x}^{\rho}} \frac{\partial}{\partial \tilde{x}^{\nu}} \frac{\partial x^{\rho}}{\partial \tilde{x}^{\mu}} (\frac{\partial x^{\rho}}{\partial \tilde{x}^{\mu}} \frac{\partial \tilde{x}^{\mu}}{\partial x^{\alpha}})V^{\alpha} - \frac{\partial \tilde{x}^{\lambda}}{\partial x^{\rho}} \frac{\partial x^{\rho}}{\partial \tilde{x}^{\mu}} \frac{\partial^{2} \tilde{x}^{\mu}}{\partial \tilde{x}^{\nu} \partial x^{\alpha}}V^{\alpha}$
Where I substituted in the above definition for the transformed connection coefficient and vector. Then, I applied the reverse chain rule and reverse product rule upon such that that te first term looks like the correct transformation of a [1, 1] tensor, meanwhile I think the second term vanishes as the term in the brackets which the derivative acts on is a kronecker delta. When it comes to the third term, I was hoping that it would come out to be the negative of the second derivative term in $\tilde{\partial}_{\mu} \tilde{V}^{\nu}$, so that overall they would cancel and we would be left with:
$$\tilde{\nabla}_{\nu}\tilde{V}^{\lambda} = \frac{\partial x^{\rho}}{\partial \tilde{x}^{\nu}} \frac{\partial \tilde{x}^{\lambda}}{\partial x^{\sigma}} \partial_{\rho}V^{\sigma} + \frac{\partial \tilde{x}^{\lambda}}{\partial x^{\rho}} \frac{\partial {x}^{\sigma}}{\partial \tilde{x}^{\nu}}\Gamma^{\rho}_{\alpha \sigma} V^{\alpha}$$
Could someone please help me with how to do this, or where I'm going wrong? Thanks :)
I would turn this question on its head, and think of defining the connection coefficients so that $\nabla_{\nu} V^{\lambda}$ transforms as a (1,1) tensor. Then I would use $\Gamma ^{\lambda}_{\mu \nu} V^{\mu} = \nabla_{\nu} V^{\lambda} - \partial_{\nu}V^{\lambda}$, and calculate the transformation law for the connection coefficients from the known (1,1) transformation law and the transformation law for $\partial_{\nu}V^{\lambda}$, calculated using the product rule as you have done in your question. If you do the calculation this way it should then be straightforward to see that the second derivative terms cancel, because you have will have in fact chosen them so that they cancel, as their defining property.
In the question you have asked it seems that the connection coefficients have been defined by their own (potentially unintuitive) transformation law. It's fine to do things this way, but I think it gives less intuition as to why they transform in this way, which probably doesn't help in doing the calculation.