Simple question about differentiation of trigonometric function

[lioric]

Explain to me:

Why the 2πf came in front?

I lost touch and sort of forgot.

 

[soarce]

This is the derivation differentiation chain rule: $$\frac{df(g(x))}{dx}=\frac{df}{dg}\cdot\frac{dg}{dx}$$
Edited by mentor...

 

[lioric]

could you please show me how this works with my example

 

[soarce]

In your case $x=t$, $f(x)=cos(x)$, $g(x)=2\pi f x$ and $f(g(x))=(f\circ g)(x)=cos(2\pi f x)$.

 

[jedishrfu]

Another way to demonstrate it is:

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

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

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

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

 

[mathman]

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