[This](https://en.wikipedia.org/wiki/Riemann_curvature_tensor) Wikipedia article explains how the Riemann curvature tensor is a measure of the failure for a tangent vector to parallel translate back to itself along an infinitesimally small loop. The article gives $X,Y,Z\in\Gamma(TM)$ and lets $tX$ and $tY$ denote the integral curves of $X$ and $Y$ respectively. They let $$\tau\_{tX}:T\_pM\to T\_pM \ \ \ \ \ \ \ \text{ and } \ \ \ \ \ \ \ \tau\_{tY}:T\_pM\to T\_pM$$ denote the parallel transport maps along the respective integral curves.

I am hoping someone can explain the computation giving this equality:

$$\left.\frac{d}{ds}\frac{d}{dt}\right|\_{t,s=0}\tau\_{sX}^{-1}\tau\_{tY}^{-1}\tau\_{sX}\tau\_{tY}Z=\left(\nabla\_X\nabla\_Y-\nabla\_Y\nabla\_X-\nabla\_{[X,Y]}\right)Z.$$

I understand that if $P\_\gamma$ denotes parallel transport along a curve $\gamma(t)$ then $$\nabla\_{\gamma^\prime(t)}Z=\left.\frac{d}{dt}\right|\_{t=0}P\_{\gamma(t)}Z$$but there is some crazy chain rule stuff going on in the curvature expression that I can't seem to get right.