How do you differentiate #g(x) = (3x-2)(2x^2-4)# using the product rule?

1 Answer
Mar 11, 2018

Answer:

#g'(x) = 18x^2 - 8x - 12#

Explanation:

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.