matrix differentiation of the first order involving differentiation of inverse of matrix by itself

Question

can some one provide proof of this differentiation \begin{equation} \frac{\partial a'Xb}{\partial X}=ab' \end{equation}
where $a'$ is transpose of matrix $a$

Answer 1

If you write the function in terms of the Frobenius Inner Product, then finding the differential and gradient is simple. $$\eqalign{ f &= a^TXb \cr &= ab^T : X \cr\cr df &= ab^T:dX \cr\cr \frac{\partial f}{\partial X} &= ab^T \cr }$$