Given y=sin x/sec(3x+1), find dy/dx

Question

It is a derivative question using Chain Rule.

Answer 1

y = Sinx/Sec(3x+1)

Apply derivative to each side with respect of x.

dy/dx(u/v) = (vu'-uv')/v^2

u = Sinx, v = Sec(3x+1)

dy/dx = [(Sec(3x+1)d/dx(Sinx)-Sinx(d/dx(Sec(3x+1))]/Sec^2(3x+1)

dy/dx = [(Sec(3x+1)Cosx-SinxSec(3x+1)Tanx(3x+1)*3]/Sec^2(3x+1)

dy/dx = [Sec(3x+1)Cosx-3SinxSec(3x+1)Tanx(3x+1)]/Sec^2(3x+1)

dy/dx = Sec(3x+1)[Cosx-3SinxTanx(3x+1)]Sec^2(3x+1)

dy/dx =[Cosx-3SinxTanx(3x+1)]Sec(3x+1)