A basic math question about the current in a capacitor

[Adir_Sh]

The current in a capacitor is calculated from

i_{C}(t)=C \dfrac{dv_{C}(t)}{dt}​

when i_{C}(t) is the current in the capacitor, t is the time variable, C is the capacitance and v_{C}(t) is the capacitor's voltage.
This equation can also be expressed, using \omega t as the variable, as

i_{C}(\omega t)=\omega C \dfrac{dv_{C}(\omega t)}{d(\omega t)}​

Now my basic question is why there's a multiplication in \omega? That is, why is it correct to multiply when changing in independent variable and multiplying in by frequency \omega? Seemingly, the new variable, \omega t, is derived in the same manner as t is derived when the function uses only t as the independent variable.
I don't quite understand this little issue. Hope someone could explain that in a sentence.

Thanks in advance!

 

[tiny-tim]

Hi Adir_Sh!

Now my basic question is why there's a multiplication in \omega?

chain rule …

dv_C/dt

= dv_C/d(ωt)*d(ωt)/dt

= dv_C/d(ωt)*ω ​

 

[rude man]

Realize that d(wt) = w dt and you have your answer.