How do you find the derivative of #y=2x(3x-1)(4-2x)#?

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Answer 1 · 2018-05-27T03:29:43

#(dy)/(dx)=-12x^3+4x^2+28x-8#

#y=2x(3x-1)(4-2x)#

#y=(6x^2-2x)(4-2x)#

#(dy)/(dx)=(6x^2-2x)(-2x)+(4-2x)(12x-2)#

#(dy)/(dx)=-12x^3+4x^2+48x-8-24x+4x#

#(dy)/(dx)=-12x^3+4x^2+28x-8#

The product rule is given by

#y=uv#

#(dy)/(dx)=uv'+vu'#