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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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    Thanks, must be more careful next time..2011-11-18

1 Answers 1

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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.2011-11-18