SOLUTION: the derivative of y=(2=3cos[2x+pi/3]/sin[x])^3

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Question 133912: the derivative of y=(2=3cos[2x+pi/3]/sin[x])^3
Answer by algebrapro18(249) About Me  (Show Source):
You can put this solution on YOUR website!
okay I am not really sure what the problem is but I think what you mean is y = %28%282%2B3cos%282x%2Bpi%2F3%29%29%2Fsin%28x%29%29%5E3

this is a VERY lengthy problem but I will try my best to fit it in here. First off there are a few rules that we need to use:

1) Quotient rule
2) Chain rule
3) cos(x+y) = cos(x) * cos(y) - sin(x)* sin(y)
4) cos%28pi%2F3%29 = 1/2
5) sin%28pi%2F3%29 = sqrt%283%29%2F2
6) Product rule

It is also better to do this problem in pieces so thats how I am going to do it, hopefully you can follow what I did.

well I think the first thing we need to do is simplify that nasty cosine expression.

cos%282x%2Bpi%2F3%29 --> Using rule number 3 from above.
= %28cos%282x%29+%2A+cos%28pi%2F3%29%29+-+%28sin%282x%29+%2A+sin%28pi%2F3%29%29 --> using 4 and 5 from above.
= %28cos%282x%29+%2A+%281%2F2%29%29+-+%28sin%282x%29+%2A+sqrt%283%29%2F2%29 --> factoring out a 1/2
= %281%2F2%29%28cos%282x%29-%28sqrt%283%29%2Asin%282x%29%29%29

now we will call this u for simplicity sake(and I don't feel like typing this over and over again). Lets find u' now because I am sure we will be using it later.

u' = (d/dx)(%281%2F2%29%28cos%282x%29-%28sqrt%283%29%2Asin%282x%29%29%29)--> By the chain rule
= %281%2F2%29%28%28-sin%282x%29%2A2%29+-%282sqrt%283%29%2Acos%282x%29%29%29 --> Multiplying by 1/2
=-sin%282x%29+-sqrt%283%29%2Acos%282x%29

Now the origonal function rewritten with u inside of it is going to be:

y= %28%282%2B3u%29%2Fsin%28x%29%29%5E3

now lets take the derivative of y

y= %28%282%2B3u%29%2Fsin%28x%29%29%5E3 --> Applying the chain rule
y' = --> apply the quotient rule
y' = --> plug in for you and du
y' =
from here its a bunch of algebra to get your answer reduced but all the calculus is done.