I am trying to take the derivative of the following:
$$\frac{\partial}{\partial \,a} \left[-\frac{1}{2} \sum \limits_{i=1}^{n} \mbox{tr} \left\lbrace \boldsymbol{\Sigma}^{-1}_{p\times p} \left(\mathbf{Y}_{p\times d}-\mathbf{M}_{p\times d} \right) \left( a^2\mathbf{A}_{d \times d} - a \mathbf{B}_{d\times d}+\mathbf{I}_{d\times d}\right) \left(\mathbf{Y}_{p\times d}-\mathbf{M}_{p\times d} \right)^T \right\rbrace \right]$$
I have done some search on the internet but have not been able to find a way of taking the derivative w.r.t. a scalar inside the trace.
I looked onto this page, and almost all formulas are for matrices.
$$-\frac{1}{2} \sum \limits_{i=1}^{n} tr \left\lbrace \boldsymbol{\Sigma}^{-1}_{p\times p} \left(\mathbf{Y}_{p\times d}-\mathbf{M}_{p\times d} \right) \left( a^2\mathbf{A}_{d \times d} - a \mathbf{B}_{d\times d}+\mathbf{I}_{d\times d}\right) \left(\mathbf{Y}_{p\times d}-\mathbf{M}_{p\times d} \right)^T \right\rbrace \\= \left[-\frac{1}{2} \sum \limits_{i=1}^{n} tr \left\lbrace \boldsymbol{\Sigma}^{-1}_{p\times p} \left(\mathbf{Y}_{p\times d}-\mathbf{M}_{p\times d} \right) \mathbf{A}_{d \times d} \left(\mathbf{Y}_{p\times d}-\mathbf{M}_{p\times d} \right)^T \right\rbrace \right]a^2 - \left[-\frac{1}{2} \sum \limits_{i=1}^{n} tr \left\lbrace \boldsymbol{\Sigma}^{-1}_{p\times p} \left(\mathbf{Y}_{p\times d}-\mathbf{M}_{p\times d} \right) \mathbf{B}_{d \times d} \left(\mathbf{Y}_{p\times d}-\mathbf{M}_{p\times d} \right)^T \right\rbrace \right]a + \left[-\frac{1}{2} \sum \limits_{i=1}^{n} tr \left\lbrace \boldsymbol{\Sigma}^{-1}_{p\times p} \left(\mathbf{Y}_{p\times d}-\mathbf{M}_{p\times d} \right) \mathbf{I}_{d \times d} \left(\mathbf{Y}_{p\times d}-\mathbf{M}_{p\times d} \right)^T \right\rbrace \right] $$ now apply formula for differentiating $f(a)=Aa^2 + Ba + C$.