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Treg. Defrentiation

  1. Dec 6, 2009 #1
    I got 2 questions

    First:
    y= Cos3(5x2-6)

    Solution :
    Y' = 3cos2(5x2-6)(-sin(5x2-6))(10x)
    y'= 30xCos2(5x2-6)(-sin(5x2-6))

    Is the correct ?


    Second
    Y=3sin4(2-x)-1

    I don't understand what do i have to do with the -1 (the power to the bracket )
    so i add it to the 4 (the power of sin ) ?

    am not sure


    Any help is appreciated
     
  2. jcsd
  3. Dec 6, 2009 #2

    rl.bhat

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    Homework Helper

    (2-x)^-1 = 1/(2-x) = u. So y = 3*sin^4(u)
    Now what is the derivative of 1/(2-x)?
     
  4. Dec 6, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, that is correct.


    That's a very peculiar notation. It would be better with an additional pair of parentheses:
    [itex] Y= 3 sin^2((2-x)^{-1})[/itex]
    That is, it is the (2- x) that is taken to the -1 power, not "sin".
     
  5. Dec 6, 2009 #4
    Thankv you guys

    So it will be

    y' = 12 sin3[(2-x)-1] Cos[(2-x)-1](-1)(2-x)-2(-1)

    right?
     
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