Simple question about differentiation of trigonometric function

• B

Main Question or Discussion Point

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 $f(g(x))=(f\circ g)(x)=cos(2\pi f x)$.

jedishrfu
Mentor
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).