Question

cos 3x find derivatives ​

Answer 1

Answer:

Prove the following identity: cos 3x = 4 cos^3x - 3cosx Here is what I did and where I'm stuck: cos 3x = cos (2x+x) = (cos 2x)(cos x) - (sin 2x)(sin x) = (2 cos^2x-1)(cos x) - [(2 sinx)(cosx)](sinx) I'm not sure how you distribute this.

Answer 2

Answer:

Step-by-step explanation:

using chain rule

dy/dx=3sin(3x)

if the question is cubed then

$\frac{dy}{dx}=cosx.cosx.(-sinx)+cosx.(-sinx)cosx+(-sinx)cosx.cosx$

we alternately differentiate the terms in multiply which are all cosx in this case

=>$\frac{dy}{dx}=-3sinx.cos^2x$