Simple question about differentiation of trigonometric function

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SUMMARY

The discussion focuses on the differentiation of the trigonometric function cos(2πft) using the chain rule. The derivation illustrates how the factor 2πf arises during differentiation, specifically in the expression d/dt(cos(2πft)). The final result confirms that the derivative is -2πf sin(2πft), demonstrating the application of the chain rule effectively. This example serves as a clear illustration of the relationship between trigonometric functions and their derivatives in the context of frequency.

PREREQUISITES
  • Understanding of the chain rule in calculus
  • Familiarity with trigonometric functions, specifically cosine
  • Basic knowledge of differentiation techniques
  • Concept of frequency in mathematical terms
NEXT STEPS
  • Study the application of the chain rule in more complex functions
  • Explore the differentiation of other trigonometric functions, such as sine and tangent
  • Learn about the implications of frequency in wave functions
  • Investigate the relationship between trigonometric functions and their graphical representations
USEFUL FOR

Students of calculus, mathematics educators, and anyone interested in understanding the differentiation of trigonometric functions in relation to frequency.

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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