\frac{\partial \ln p}{\partial \boldsymbol{\Sigma}_k^{-1}} = \sum_{n=1}^N \frac{\pi_k \mathcal{N}(\mathbf{x}_n|\boldsymbol{\mu}_k, \boldsymbol{\Sigma}_k)}{\sum_{j=1}^K \pi_j \mathcal{N}(\mathbf{x}_n|\boldsymbol{\mu}_j, \boldsymbol{\Sigma}_j)} \cdot \frac{\partial}{\partial \boldsymbol{\Sigma}_k^{-1}} \ln \mathcal{N}(\mathbf{x}_n|\boldsymbol{\mu}_k, \boldsymbol{\Sigma}_k)