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$A(t)= \pmatrix{\alpha(t)^2&2\alpha(t)\beta(t)&...&...\\...&\alpha(t)\bar{\alpha}(t)-\beta(t)\bar{\beta}(t)&...&...\\...&...&2\alpha\bar{\alpha}\beta-\beta^2\bar{\beta}&...}$

I am meant to find $A'(t)$ and thus $A'(0)$. The $...$ are just other terms that I did not write out as its not necessary for my question below (some are quite long!).

So obviously for $A'(t)$, terms like $\alpha(t)^2$ become $2\alpha'(t)$, and I used the product rule to differentiate $2\alpha(t)\beta(t)$ and it becomes $2(\alpha'(t)\beta(t)+\alpha(t)\beta'(t))$ etc..

The problem is when I come across $2\alpha\bar{\alpha}\beta-\beta^2\bar{\beta}$, how would I find $A'(t)$? Since there are 3 terms $\alpha$, $\bar{\alpha}$ and $\beta$.

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The derivative of $\alpha^2$ is not $2\alpha'$ but $2\alpha\alpha'$. Don't forget the chain rule! – Gerry Myerson Nov 17 '11 at 23:55
Thanks, must be more careful next time.. – Ray Nov 18 '11 at 0:16
up vote 3 down vote accepted

If you have the product of three terms, say $\alpha\beta\gamma$, consider it as the product of two terms, $\alpha\beta$ and $\gamma$. Differentiate according to the product rule, which will require differentiating $\gamma$ and $\alpha\beta$ separately, the latter requiring the product rule itself. In essence, the rule works recursively as many times as required to isolate individual terms.

Then, $(\alpha\beta\gamma)'$ becomes $(\alpha\beta)'\gamma + (\alpha\beta)\gamma'$ which is $(\alpha'\beta + \alpha\beta')\gamma + (\alpha\beta)\gamma'$. Of course that can be simplified.

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Thanks for your help, understand it. – Ray Nov 18 '11 at 0:16

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