Question on the Palatini identity for the metric variation of the Riemann tensor

Question

Let $({X},\mathcal{O},\mathcal{A},{g},\nabla)$ be a smooth ${n}$-dimensional manifold with Riemannian metric ${g}$ and the Levi-Civita connection ${\nabla}$. We now define an action functional that is dependent on the metric field \begin{align*} {W}[g]=\int_{X}\mathcal{L}(x)[{g}_{\mu\nu}(x)]{\,}\mathrm{d}^{n}{x}=\int_{X}{R}(x)\sqrt{\mathrm{det}({g}_{\mu\nu}(x))}{\,}\mathrm{d}^{n}{x}{\,}{,} \end{align*} where ${R}=\mathrm{Ric}_{\mu\nu}{g}^{\mu\nu}$ is the Ricci curvature. After variation of said functional one arrives at \begin{align*} \frac{\delta{W}[g]}{\delta{g}^{\pi\tau}(y)}=\int_{X}\bigg(\frac{\delta{R}(x)}{\delta{g}^{\pi\tau}(y)}+\frac{{R}(x)}{\sqrt{\mathrm{det}({g}_{\mu\nu}(x))}}\frac{\delta\sqrt{\mathrm{det}({g}_{\mu\nu}(x))}}{\delta{g}^{\pi\tau}(y)}\bigg)\sqrt{\mathrm{det}({g}_{\mu\nu}(x))}{\,}\mathrm{d}^{n}{x}{\,}{.} \end{align*} To now calculate the variational derivative of the Ricci scalar with respect to the metric, one would have to first calculate for the Riemann tensor \begin{align*} \frac{\delta{\mathrm{Riem}^{\omega}}_{\alpha\mu\nu}(x)}{\delta{g}^{\pi\tau}(y)}&=\frac{\partial}{\partial{x}^{\mu}}\bigg[\frac{\delta\Gamma^{\omega}_{\alpha\nu}(x)}{\delta{g}^{\pi\tau}(y)}\bigg]-\frac{\partial}{\partial{x}^{\nu}}\bigg[\frac{\delta\Gamma^{\omega}_{\mu\alpha}(x)}{\delta{g}^{\pi\tau}(y)}\bigg]+\Gamma^{\omega}_{\mu\sigma}(x)\frac{\delta\Gamma^{\sigma}_{\alpha\nu}(x)}{\delta{g}^{\pi\tau}(y)}+\Gamma^{\sigma}_{\alpha\nu}(x)\frac{\delta\Gamma^{\omega}_{\mu\sigma}(x)}{\delta{g}^{\pi\tau}(y)}\\ &-\Gamma^{\omega}_{\sigma\nu}(x)\frac{\delta\Gamma^{\sigma}_{\mu\alpha}(x)}{\delta{g}^{\pi\tau}(y)}-\Gamma^{\sigma}_{\mu\alpha}(x)\frac{\delta\Gamma^{\omega}_{\sigma\nu}(x)}{\delta{g}^{\pi\tau}(y)}{\,}{.} \end{align*} The Palatini identity now holds \begin{align*} \frac{\delta{\mathrm{Riem}^{\omega}}_{\alpha\mu\nu}(x)}{\delta{g}^{\pi\tau}(y)}=\frac{\mathrm{D}}{\mathrm{D}{x}^{\mu}}\bigg[\frac{\delta\Gamma^{\omega}_{\alpha\nu}(x)}{\delta{g}^{\pi\tau}(y)}\bigg]-\frac{\mathrm{D}}{\mathrm{D}{x}^{\nu}}\bigg[\frac{\delta\Gamma^{\omega}_{\mu\alpha}(x)}{\delta{g}^{\pi\tau}(y)}\bigg]{\,}{,} \end{align*} where the operator $\frac{\mathrm{D}}{\mathrm{D}{x}^{\mu}}$ represents the covariant derivative. However when I now calculate the covariant derivative of the metric variation of the connection coefficients i arrive at \begin{align*} \frac{\mathrm{D}}{\mathrm{D}{x}^{\mu}}\bigg[\frac{\delta\Gamma^{\omega}_{\alpha\nu}(x)}{\delta{g}^{\pi\tau}(y)}\bigg]&=\frac{\partial}{\partial{x}^{\mu}}\bigg[\frac{\delta\Gamma^{\omega}_{\alpha\nu}(x)}{\delta{g}^{\pi\tau}(y)}\bigg]+\Gamma^{\omega}_{\mu\sigma}(x)\frac{\delta\Gamma^{\sigma}_{\alpha\nu}(x)}{\delta{g}^{\pi\tau}(y)}-\Gamma^{\sigma}_{\mu\alpha}(x)\frac{\delta\Gamma^{\omega}_{\sigma\nu}(x)}{\delta{g}^{\pi\tau}(y)}-\Gamma^{\sigma}_{\mu\nu}(x)\frac{\delta\Gamma^{\omega}_{\alpha\sigma}(x)}{\delta{g}^{\pi\tau}(y)}\\ &-\Gamma^{\sigma}_{\mu\pi}(x)\frac{\delta\Gamma^{\omega}_{\alpha\nu}(x)}{\delta{g}^{\sigma\tau}(y)}-\Gamma^{\sigma}_{\mu\tau}(x)\frac{\delta\Gamma^{\omega}_{\alpha\nu}(x)}{\delta{g}^{\pi\sigma}(y)}{\,}{.} \end{align*} For this to eventually yield the Palatini identity, those last two summands would have to vanish, but im not sure on how to argue for this. My question is now if they are each simply $0$ or if at least the the term \begin{align*} \Gamma^{\sigma}_{\pi\nu}(x)\frac{\delta\Gamma^{\omega}_{\mu\alpha}(x)}{\delta{g}^{\sigma\tau}(y)}+\Gamma^{\sigma}_{\tau\nu}(x)\frac{\delta\Gamma^{\omega}_{\mu\alpha}(x)}{\delta{g}^{\pi\sigma}(y)}-\Gamma^{\sigma}_{\mu\pi}(x)\frac{\delta\Gamma^{\omega}_{\alpha\nu}(x)}{\delta{g}^{\sigma\tau}(y)}-\Gamma^{\sigma}_{\mu\tau}(x)\frac{\delta\Gamma^{\omega}_{\alpha\nu}(x)}{\delta{g}^{\pi\sigma}(y)} \end{align*} cancels out?