Posted by T.Gracken on October 21, 2002 at 19:17:22:
In Reply to: Re: Derivatives of Trig Functions posted by mfrizza on October 21, 2002 at 18:38:30:
: Thanks, I definitely see what to do now. I do have a quick follow-up regarding...
: y' = [-2sin(x) - sin^2(x) - cos^2(x)]/[4 + 4sin(x) + sin^2(x)]
: you factored out a -1 of the numerator. I'm confused as to what happened with the -2sinx. If you factor out a -1, wouldn't it be 2sinx + sin^2x + cos^2x / denom?
yes! (sorry about that), so you have -[2sin(x)+1]/denom
...and that will screw up the rest of what I wrote. instead, you should solve: -[2sin(x)+1]=0
which will lead to sin(x)=-1/2
giving x = 7pi/6 + 2pi*n or x = 11pi/6 + 2pi*n (n an integer)
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