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B Simple question about differentiation of trigonometric function

  1. May 23, 2016 #1
    20160523_143427.jpg
    Explain to me:

    Why the 2πf came in front?

    I lost touch and sort of forgot.
     
    Last edited by a moderator: May 23, 2016
  2. jcsd
  3. May 23, 2016 #2
    This is the derivation differentiation chain rule: [tex]\frac{df(g(x))}{dx}=\frac{df}{dg}\cdot\frac{dg}{dx}[/tex]
    Edited by mentor...
     
    Last edited by a moderator: May 23, 2016
  4. May 23, 2016 #3
    could you please show me how this works with my example
     
  5. May 23, 2016 #4
    In your case ##x=t##, ##f(x)=cos(x)##, ##g(x)=2\pi f x## and ##f(g(x))=(f\circ g)(x)=cos(2\pi f x)##.
     
  6. May 23, 2016 #5

    jedishrfu

    Staff: Mentor

    Another way to demonstrate it is:

    ##d/dt ( cos (2 * \pi * f * t) ) = ##

    ## = - sin ( 2 * \pi * f * t ) * d/dt ( 2 * \pi * f * t ) ##

    ## = - sin ( 2 * \pi * f * t ) * ( 2 * \pi * f ) ##

    ## = - ( 2 * \pi * f * t ) * sin ( 2 * \pi * f * t ) ##
     
  7. May 23, 2016 #6

    mathman

    User Avatar
    Science Advisor
    Gold Member

    General formula. Let f'(x)=g(x), then f'(ax)=ag(ax).
     
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