May 23, 2016 #1 lioric Messages 335 Reaction score 26 Explain to me: Why the 2πf came in front? I lost touch and sort of forgot. Last edited by a moderator: May 23, 2016
May 23, 2016 #2 soarce Messages 154 Reaction score 24 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: May 23, 2016
This is the derivation differentiation chain rule: \frac{df(g(x))}{dx}=\frac{df}{dg}\cdot\frac{dg}{dx} Edited by mentor...
May 23, 2016 #3 lioric Messages 335 Reaction score 26 could you please show me how this works with my example
May 23, 2016 #4 soarce Messages 154 Reaction score 24 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)##.
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)##.
May 23, 2016 #5 jedishrfu Mentor Insights Author Messages 15,468 Reaction score 10,177 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 ) ##
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 ) ##
May 23, 2016 #6 mathman Science Advisor Homework Helper Messages 8,130 Reaction score 573 General formula. Let f'(x)=g(x), then f'(ax)=ag(ax).