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Derivative of trigonometric functions

  1. Mar 1, 2013 #1
    1. The problem statement, all variables and given/known data

    g(x) = 4∏ [cos(3∏x) sin (3∏x)]


    3. The attempt at a solution

    g(x) = 4∏ [cos(3∏x) sin (3∏x)]'

    4∏{[cos (3∏x)][sin(3∏x)]' + [sin(3∏x)][cos(3∏x)]'} =

    4∏{[cos (3∏x)][cos(3∏x) . (3∏)] + [sin(3∏x)][-sin(3∏x) . (3∏)] =

    Now, my question is: Can I combine the numbers and have the answer as:

    - 36∏ [cos2(3∏x) + sin2(3∏x)]

    Thank you so much!!
     
  2. jcsd
  3. Mar 1, 2013 #2

    Dick

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    The -36 is certainly wrong and isn't it a difference between the cos^2 and sin^2 parts? Why would you think you can do that? Show how you combined. Use algebra.
     
  4. Mar 1, 2013 #3
    Oh yes. I was thinking of transferring the negative sign from sin to 3∏.

    How about this final answer:

    -12∏ sin2(3∏x) + 12∏cos2(3∏x)
     
  5. Mar 1, 2013 #4

    Dick

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    That's much better.
     
  6. Mar 1, 2013 #5
    Thank you SO MUCH!
     
    Last edited: Mar 1, 2013
  7. Mar 2, 2013 #6
    Oh I have a question:

    In the following problem

    5 sin (8∏x)

    They isolated the 5 to then find the derivative of sin (8∏x)

    meaning they are not taking the derivative of 5, right? because taking its derivative would

    result in zero.

    Now, for the problem

    cos (sec (5∏x))

    I believe I am supposed to take the derivate of cos.

    How do I know when I am supposed to take the derivative of these numbers?
     
  8. Mar 2, 2013 #7

    Dick

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    You are probably overcomplicating this. Taking the the derivative of numbers is never a problem, the derivatives of them are zero. The real problem is with the parts that are functions of x. Your last problem needs to use the chain rule. Look it up if you don't know it.
     
  9. Mar 2, 2013 #8

    HallsofIvy

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