Findin extrema of function ???

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I know this is a very simple problem, but for some reason i have a block...

the question asks for the extrema of the following equation:

theta(t) = arccos ((t^2+1) / (sqrt (t^2+1) * sqrt(4t^2+1)))

I know to find the extrema you must first take the derivative. I tried plugging it into my TI89 Titanium but it says too many arguments, and i have NO idea how to do it by hand.

Secondly, i know i must set this derivative equal to 0 and find what t would have to be. But how would i even do that by hand?

I really need some help on this one, does anyone know what the derivative is? What about the second derivative? Thanks!
 
Aren't you doing the same problem in:

viewtopic.php?f=3&t=29703

More and more it sounds like a take-home exam as opposed to review problem.

Anyway, use implicit differentiation, or chain rule of diffrentiation.
 
I know, but how would i do that with the arccos? My teacher said this problem is important and hinted there would be a problem just like it, so i REALLY need to know how to do it. Is arccos just cos^-1? I have no idea how to take the derivative of this type of thing.. thanks for your help
 
For all reals. Let y = arccos(t^2+1), then cos(y) = t^2+1

-1<=cos(y)<=1, -1<=t^2+1<=1, -2<=t^2<=0, sqrt(-2)<=|t|<=0, but |t|>=0, hence t can only = 0

Ergo theta(t) = arccos(1)/1 = 0/1 =0, hence theta ' (t) = 0 for all t.

theta(t) is what is called a not well defined function.
 
yes...

\(\displaystyle arccos(t) \, = \, cos^{-1}(t)\)

so

\(\displaystyle \frac{d[arccos(t)]}{dt} \, = \, \frac{d[cos^{-1}(t)]}{dt} \, = \, - \frac{1}{\sqrt{1-t^2}}\)
 
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