Start with the product rule for 3 factors
#[fgh]'=f'gh+fg'h+fgh'#
In this case, let #f=(2xsinx)^3#, #g=x#, and #h=cosx#
#y'=color(red)([(2xsinx)^3]')xcosx+(2xsinx)^3color(gray)(x')cosx+(2xsinx)^3xcolor(purple)([cosx]')#
To find #color(red)([(2xsinx)^3]')# we need to use the chain rule
#[f(g)]'=f'(g)timesg'#
In this case, let #f=g^3# and #g=2xsinx#
#color(red)([f(g)]'=3(2xsinx)^2times)color(green)([2xsinx]')#
To find #color(green)([2xsinx]')# we need to use the product rule for 2 factors
#[fg]'=f'g+fg'#
In this case, let #f=2x# and #g=sinx#
#color(green)([fg]'=2sinx+2xcosx)#
#color(red)([f(g)]'=3(2xsinx)^2times)color(green)(2sinx+2xcosx)#
#color(red)([(2xsinx)^3]'=3(2xsinx)^2times2sinx+2xcosx)#
#color(gray)(x')# is easy to find
#color(gray)(x'=1)#
#color(purple)([cosx]')# is also something you should know
#color(purple)([cosx]'=-sinx#
#y'=color(red)([3(2xsinx)^2times2sinx+2xcosx])xcosx+(2xsinx)^3color(gray)(times1times)cosx+(2xsinx)^3xcolor(purple)((-sinx))#
Depending on your teacher, that may be as far as you need to go but some will want it simplified, which I will do below.
#y'=6x(2xsinx)^2sinxcosx+(2xsinx)^3cosx-x(2xsinx)^3sinx#
Equivalent forms:
#y'=(2xsinx)^2(6xsinxcosx+2xsinxcosx-2x^2sin^2x)#
#y'=(2xsinx)^3(3cosx+cosx-xsinx)#
#y'=(2xsinx)^3(4cosx-xsinx)#
