The derivative of the given product is determined by applying the
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product rule:
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#color(blue)((uv)'=u'v+v'u)#
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Derivative of powers is also used in this exercise:
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#color(red)(((u(x))^n)'=n(u(x))^(n-1)xxu'(x)#
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#y=(2x+1)^5(x^3-x+1)^4#
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#y'=((2x+1)^5(x^3-x+1)^4)'#
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#color(blue)(y'=((2x+1)^5)'(x^3-x+1)^4 + ((x^3-x+1)^4)'(2x+1)^5#
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#y'=color(red)(5(2x+1)^4xx(2x+1)')(x^3-x+1)^4+color(red)(4(x^3-x+1)^3(x^3-x+1)')(2x+1)^5#
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#y'=(5(2x+1)^4xx2)(x^3-x+1)^4+4(x^3-x+1)^3(3x^2-1)(2x+1)^5#
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#y'=10(2x+1)^4(x^3-x+1)^4+4(x^3-x+1)^3(3x^2-1)(2x+1)^5#
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#y'=2(2x+1)^4(x^3-x+1)^3(5(x^3-x+1)+2(3x^2-1)(2x+1))#
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#y'=2(2x+1)^4(x^3-x+1)^3(5x^3-5x+5+2(6x^3+3x^2-2x-1))#
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#y'=2(2x+1)^4(x^3-x+1)^3(5x^3-5x+5+12x^3+6x^2-4x-2)#
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#y'=2(2x+1)^4(x^3-x+1)^3(17x^3+6x^2-9x+3)#
