How do you differentiate #(x^2-1)(2x)-(x^2-1)(2x)#?

1 Answer
Andy Y. · Jim H
Aug 3, 2015

The answer is #d/dx[(x^2-1)(2x)-(x^2-1)(2x)]=0#

Explanation:

If #f(x) = (x^2-1)(2x)-(x^2-1)(2x)# and #x# exists, then
#f(x)=0#.

Luckily the derivative of a constant is zero, so

#d/dx [(x^2-1)(2x)-(x^2-1)(2x)] = d/dx[0]=0.#

You could also practice your rules of derivatives if you like, so that

#d/dx[(x^2-1)(2x)-(x^2-1)(2x)]#

By the Product Rule
#=d/dx[(x^2-1)(2x)]-d/dx[(x^2-1)(2x)]#
#=[(x^2-1)2+2x(2x)]-[(x^2-1)2+2x(2x)]#
#=[2x^2-2+4x^2]-[2x^2-2+4x^2]#
#=[6x^2-2]-[6x^2-2]#
#=(6x^2-6x^2)+(2-2)#
#=0+0=0#

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