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Derivative of trig functions

  1. Mar 2, 2013 #1
    1. The problem statement, all variables and given/known data

    x3 - sin 2x

    Find f'(∏/6)


    3. The attempt at a solution

    f'(x) = 3x2 - 2 cos 2x
    f(∏/6) = 2700 - 2 [ (√3/2) ] ---> from 2 [ cos(∏/6)]

    answer: 2700 - √3

    My book has the answer as (∏2 - 12)/12
     
  2. jcsd
  3. Mar 2, 2013 #2
    Work in radians so...........

    3(pi^2/36) - 2cos2(pi/6)

    (pi^2/12) - 1 = (pi^2-12)/12

    I hope I helped :S
     
  4. Mar 2, 2013 #3
    But is it wrong if I convert ∏/6 = 30 and work with that?
     
  5. Mar 2, 2013 #4
    Umm when I put (pi/6) into the gradient function 3x^2 - 2 cos 2x

    I get 3(pi/6)^2 which is 3 (pi^2/36) minus 2cos2(pi/6)

    2 times pi/6 is pi/3 and the cos of pi/3 is 1/2 and 1/2 times 2 is 1.

    so the whole thing ends up being (pi^2/12)- 1
    1= 12/12
    so the last line is (pi^2)-(12)/12
     
  6. Mar 2, 2013 #5
    But isn't (∏)^2/12 = 2700 and 12/12 = 1

    Wouldn't it be sort of equivalent to 2700 - sqrt of 3 (1.73) ?

    That's what I am trying to understand. What's the difference if I work with 30 instead of ∏/6 ? Is it wrong? =/
     
    Last edited: Mar 2, 2013
  7. Mar 2, 2013 #6
    Wait.. how did you get 2700?
     
  8. Mar 2, 2013 #7
    Oh I was thinking of ∏ = 180.
    Is it wrong? =/
     
  9. Mar 2, 2013 #8
    ohhhhh lol yes it's wrong don't think of it like that. In questions
    you're either working in radians or degrees . In this case it is radians.

    So so pi/6 is not 180/6 , if it was degrees then it would be 30 degrees.
     
  10. Mar 2, 2013 #9
    It is very, very wrong. Pi radians is 180 degrees yes. But the number pi is not equal to the number 180.
     
  11. Mar 2, 2013 #10
    nr. pi = 3.14, right?
    Good you clarified that. Now I wont make the same mistake :)
    180 and 3.14 were sort of mixed in mind.
    Thank you so much! =D
     
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