Could someone explain this equation?
$$ \frac{d \operatorname{tr}(AXB)}{d X} = BA $$
I understand that
$$ d\operatorname{tr}(AXB) = \operatorname{tr}(BA \; dX) $$
but I don't quite understand how to move $dX$ out of the trace.
Could someone explain this equation?
$$ \frac{d \operatorname{tr}(AXB)}{d X} = BA $$
I understand that
$$ d\operatorname{tr}(AXB) = \operatorname{tr}(BA \; dX) $$
but I don't quite understand how to move $dX$ out of the trace.