Simple question about differentiation of trigonometric function

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Discussion Overview

The discussion centers around the differentiation of trigonometric functions, specifically focusing on the appearance of the term 2πf in the context of differentiating the cosine function. Participants explore the application of the chain rule in this differentiation process.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant asks for clarification on why the term 2πf appears in front of the differentiation of a trigonometric function.
  • Another participant presents the chain rule for differentiation, indicating how it applies to the function in question.
  • A request is made for a specific example to illustrate the differentiation process.
  • In response, a participant provides a specific case using the cosine function and the term 2πf, detailing the composition of functions involved.
  • Another participant demonstrates the differentiation step-by-step, showing how the derivative of cos(2πft) is calculated using the chain rule.
  • A general formula for differentiation is mentioned, suggesting a broader application of the concepts discussed.

Areas of Agreement / Disagreement

Participants appear to be engaged in a constructive discussion with no clear consensus or disagreement noted. The conversation remains focused on clarifying the differentiation process without resolving any potential misunderstandings.

Contextual Notes

Some assumptions about the familiarity with differentiation and trigonometric functions may be present, as well as the dependence on the definitions of the functions involved.

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...
 
Last edited by a moderator:
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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