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Is the differentiation of the law of cosines ($c^2= a^2 + b^2 - 2ab\cos C$) this? a, b, c, A, B, and C are functions of time t.

$$2c \frac{dc}{dt} = 2a \frac{da}{dt} + 2b\frac{db}{dt} - 2b \cos C \frac{da}{dt} - 2 a \cos C \frac{db}{dt} + 2ab \sin C \frac{dC}{dt}$$

Josue
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    An identity has no derivative; can you please state your problem more clearly? – egreg Sep 12 '14 at 14:58
  • Sorry, egreg! I meant the differentiation. – Josue Sep 12 '14 at 15:04
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    It doesn't make much more difference. – egreg Sep 12 '14 at 15:04
  • I don't know how to answer that, then. I know I'm supposed to differentiate both sides of the equation of the law of cosines using the Chain Rule. Then I put in the given and found values of the problem I'm doing and solve for dc/dt. I think I found my error, anyway. I confused DC/dt with dc/dt. Thank you, anyway. – Josue Sep 12 '14 at 15:10
  • @egreg Well, if $a,b,c,A,B,C$ are functions of $t$, it's possible to differentiate. Maybe not very useful, but possible. It just amounts to write that if two functions (of $t$) are equal, then their derivatives are also equal. – Jean-Claude Arbaut Sep 12 '14 at 15:10
  • @Josue What are you trying to do exactly? Why did you want to differentiate this in the first place? – Jean-Claude Arbaut Sep 12 '14 at 15:12
  • I just wanted to verify that my differentiation of both sides of the equation was correct. I couldn't get the correct answer to the problem, and I thought that maybe it was because I didn't differentiate both side of the equation correctly. I just realized that the reason for not getting the correct answer was that I confused dc/dt with dC/dt. I just got the right answer. Thank you very much to all of you. – Josue Sep 12 '14 at 15:14
  • It was for a dfferemtiation application problem. I finally figure out why I was not getting the answer. Thank you, Jean-Claude! – Josue Sep 12 '14 at 15:17
  • Travis' editing of the equation help me figure out my mistake. Thank you, Travis! – Josue Sep 12 '14 at 15:21
  • I also just realized that I forgot to mention that a, b, c, A, B, C were functions of time t. Sorry for the confusion. – Josue Sep 12 '14 at 15:25
  • And that was one very important piece of information, @Josue. Perhaps you should add that in your question. – Timbuc Sep 12 '14 at 15:26
  • Yes, Timbuc. I'm noticing how badly I expressed the question. – Josue Sep 12 '14 at 15:27
  • I just added it, Timbuc. Thanks! – Josue Sep 12 '14 at 15:30

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