We know,
The product rule says :
#d/dxf(x)g(x) = f'(x) * g(x) + f(x) * g'(x) = d/dxf(x) * g(x) + f(x) * d/dxg(x)#
When #f(x)# and #g(x)# are two differentiable polynomials.
[Don't be confused with the question function #g(x)# and the example here.]
So,
#g'(x) = d/dxg(x) = d/dx(3x - 2)(2x^2 - 4)#
#= (d/dx(3x - 2)) xx (2x^2 - 4) + (3x - 2) xx (d/dx (2x^2 - 4))#
#= (d/dx3x - d/dx 2) xx (2x^2 - 4) + (3x -2) xx (d/dx2x^2 - d/dx 4)#
[Subtraction Rule]
#= (3 - 0)(2x^2 - 4) + (3x - 2)(2* 2 x^(2 -1) - 0)# [The derivative of a constant is always zero, and #d/dx x^n = nx^(n-1)#]
#= (6x^2 - 12) + 4x(3x - 2)#
#= 6x^2 - 12 + 12x^2 -8x#
#= 18x^2 - 8x - 12#
So, Got The Answer, and I hope this will help.
