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Trigonometric differentiation

  1. Jun 19, 2005 #1
    When y=sin(pi*x), why does y'=cos(pi*x)*pi, not y'=cos(pi*x)?

  2. jcsd
  3. Jun 19, 2005 #2


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    It's the chain rule. when the argument of a trig function is a FUNCTION of x, you have take the derivative of the agrgument. So, in general

    (f(g(x)))'= f'(g(x))*g'(x)
  4. Jun 20, 2005 #3
    So than in this case y=f(u)=sinu and u=g(x)=pi*x, correct?
  5. Jun 20, 2005 #4


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    That is correct.
  6. Jun 20, 2005 #5
    Thanks guys, I appreciate it.

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