How do you differentiate #f(x)=(x^2+x)(x-1)# using the product rule?

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Answer 1 · 2018-05-23T00:04:27

#f'(x)=3x^2-1#

product rule: #d/dx(g(x)h(x))=g'(x)h(x)+h'(x)g(x)#

for this question, #g(x)=x^2+x# and #h(x)=x-1#

#d/dx((x^2+x)(x-1))=d/dx(x^2+x)(x-1)+d/dx(x-1)(x^2+x)#
#=(2x+1)(x-1)+1(x^2+x)#
#=2x^2-2x+x-1+x^2+x#
#=3x^2-1#

Answer 2 · 2018-05-23T00:06:24

#f(x) = (x^2+x)(x-1)#

#f'(x) = (x^2+x)(1) + (2x+1)(x-1)#

#f'(x) = x^2+x + (2x+1)(x-1)#

#f'(x) = x^2 + x + 2x^2 - 2x + x - 1#

#f'(x) = 3x^2 - 1#