Matrix Commutativity **Question 1** If I have a 3 by 3 skew symmetric matrix, K(t) (K(t) and K'(t) doesn't commute) what would it be $\frac{\mathrm{d}(K(t)^n) }{\mathrm{d} t} \tag 1$? is it $(n) K(t)^{n-1}K'(t) \tag 2$ **Question 2** Then what could be $\frac{\mathrm{d}(e^{K(t) })}{\mathrm{d} t}\tag 3$ Is it like the following $\frac{\mathrm{d}(e^{K(t)}) }{\mathrm{d} t} =\frac{\mathrm{d}\\{ I+K(t)+\frac{K(t)^2}{2!}+.+.+. \\} }{\mathrm{d} t} \tag 4$ $= K'(t)+\frac{K(t)K'(t) }{1!}+.+.+. =e^{K(t)}K'(t) \tag 5$ So it is obvious that $\frac{\mathrm{d}(e^{K(t) })}{\mathrm{d} t} \ne K'(t)e^{K(t)} \tag 6 $ as long as K(t) and K'(t) wont commutate. But is it correct to say $\frac{\mathrm{d}(e^{K(t) })}{\mathrm{d} t} = e^{K(t)}K'(t) \tag 7 $ NB:: I am bit confused regarding this. As per this note link, it is says commutativity property of K(t),K'(t). But it never says (7) is wrong

When multiplication is non-commutative, you have to apply the product rule for derivatives more carefully: $$\frac{d}{dt}(K(t)^2) = \frac{d}{dt}(K(t) \cdot K(t)) = K'(t) \cdot K(t) + K(t) \cdot K'(t)$$ Notice the different ordering of the 2 terms on the right side. This is _not_ equal to $2 K(t) K'(t)$ unless $K(t)$ and $K'(t)$ commute. So (2) is wrong so (7) is also wrong.