How do you differentiate #g(x) = (x^2+1) (x^2-3x)# using the product rule?

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Answer 1 · 2016-01-04T06:40:27

#g'(x)=4x^3-6x^2+2x-2#

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

Product rule :#d/dx(uv)=(du)/dxv+u(dv)/dx#

#u=(x^2+1)#

#du/dx=2x#

#v=x^2-2x#

#dv/dx=2x=2#

#d/dx(x^2+1)(x^2-2x)=(du)/dxv+u(du)/dx#

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

=#2x^3-4x^2+2x^3-2x^2+2x-2#

=#4x^3-6x^2+2x-2#