EM Algorithm - Derivative wrt covariance I am struggling trying to find the derivative of the expression below $\frac{\partial}{\partial \sum_k^{-1}} $ wrt the covariance matrix $\sum_k^{-1}$
$ \max \sum_{n=1}^N \sum_{k=1}^K q_{kn} \log ( \frac{\pi_k}{q_{kn}} \times \frac{1}{\sqrt{2\pi |\sum_k|}}e^{-\frac{1}{2}(x_n - \mu_k)^T\sum_k^{-1}(x_n - \mu_k)})  $
Below is the solution. I was hoping someone could show me the way.

 A: This answer and this tutorial give the derivatives of log multivariate Gaussians, the rest should be easy.
$$Q=\sum_n\sum_kq_{kn}[\log\pi_k-\frac{k}{2}\log 2\pi -\frac{1}{2}\log\mid\Sigma_k\mid -\frac{1}{2}(x_n-\mu_k)^T\Sigma_k^{-1}(x_n-\mu_k)]$$
$$\frac{\partial Q}{\partial\mu_k}=\sum_nq_{kn} \frac{-\frac{1}{2}\partial (x_n-\mu_k)^T\Sigma_k^{-1}(x_n-\mu_k)}{\partial\mu_k}=\sum_nq_{kn}\Sigma_k^{-1}(x_n-\mu_k)$$
$$\frac{\partial Q}{\partial\Sigma_k}=\sum_nq_{kn} \frac{-\frac{1}{2}\partial\log\mid\Sigma_k\mid -\frac{1}{2}\partial(x_n-\mu_k)^T\Sigma_k^{-1}(x_n-\mu_k)}{\partial\Sigma_k}$$$$=-\frac{1}{2}\sum_nq_{kn}(\Sigma_k^{-1}-\Sigma_k^{-1}(x_n-\mu_k)(x_n-\mu_k)^T\Sigma_k^{-1})$$
Setting the derivatives to zero we can get the desired results.
$$-\frac{1}{2}\sum_nq_{kn}(\Sigma_k^{-1}-\Sigma_k^{-1}(x_n-\mu_k)(x_n-\mu_k)^T\Sigma_k^{-1})=0$$
$$-\frac{1}{2}\sum_nq_{kn}\Sigma_k(\Sigma_k^{-1}-\Sigma_k^{-1}(x_n-\mu_k)(x_n-\mu_k)^T\Sigma_k^{-1})\Sigma_k=0$$
$$-\frac{1}{2}\sum_nq_{kn}(\Sigma_k-(x_n-\mu_k)(x_n-\mu_k)^T)=0$$
$$\sum_nq_{kn}\Sigma_k-\sum_nq_{kn}(x_n-\mu_k)(x_n-\mu_k)^T=0$$
$$\Sigma_k=\frac{\sum_nq_{kn}(x_n-\mu_k)(x_n-\mu_k)^T}{\sum_nq_{kn}}$$
