Simple question about differentiation of trigonometric function

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  • Thread starter lioric
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  • #1
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20160523_143427.jpg

Explain to me:

Why the 2πf came in front?

I lost touch and sort of forgot.
 
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  • #2
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This is the derivation differentiation chain rule: [tex]\frac{df(g(x))}{dx}=\frac{df}{dg}\cdot\frac{dg}{dx}[/tex]
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  • #3
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could you please show me how this works with my example
 
  • #4
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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)##.
 
  • #5
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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 ) ##
 
  • #6
mathman
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General formula. Let f'(x)=g(x), then f'(ax)=ag(ax).
 

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