How do you use the product rule to differentiate #y=(x+1)^2(2x-1)#?

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Answer 1 · 2018-03-26T11:19:23

So I also need to use chain rule on #(x+1)^2#

#dy/dx = u'v + v'u#
#u' = 2(x+1) * 1#
#v' = 2#

#u=(x+1)^2#
#v=(2x-1)#

subbing into the product rule.

#dy/dx = 2(2x+1) * (2x-1) + 2(x+1)^2#

#dy/dx = 2(4x^2-1) + 2(x^2+2x+1)#

#dy/dx = 8x^2-2 + 2x^2+4x+2#

#dy/dx = 10x^2+4x#

Answer 2 · 2018-03-26T11:26:31

#dy/dx=2x(x+1)^2+2(x+1)(2x-1)#
or
#dy/dx=2x^3+8x^2+4x-2#

We know that a product is to things multiplied by each other so #(x+1)^2# and #(2x-1)# are separate products

#u=(x+1)^2#
#u'=2(x+1)*1#

#v=2x-1#
#v'=2x#

The product rule is #dy/dx=uv'+vu'#

so it is

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

simplified

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

Further simplification

#dy/dx=2x^3+6x^2-2x+2x^2+6x-2#
#dy/dx=2x^3+8x^2+4x-2#