High School Simple question about differentiation of trigonometric function

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The discussion focuses on the differentiation of the function cos(2πft) using the chain rule. The term 2πf appears as a result of differentiating the inner function g(x) = 2πfx. The differentiation process shows that d/dt(cos(2πft)) involves multiplying by the derivative of the inner function, which is 2πf. The final result illustrates how the chain rule applies to this specific trigonometric function. Understanding this derivation clarifies the role of the coefficient 2πf in the differentiation process.
lioric
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Explain to me:

Why the 2πf came in front?

I lost touch and sort of forgot.
 
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This is the derivation differentiation chain rule: \frac{df(g(x))}{dx}=\frac{df}{dg}\cdot\frac{dg}{dx}
Edited by mentor...
 
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could you please show me how this works with my example
 
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)##.
 
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 ) ##
 
General formula. Let f'(x)=g(x), then f'(ax)=ag(ax).
 

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