Haven't done one of these in forever. We avoid the quotient rule with an implicit logarithmic differentiation.
[tex]y=\dfrac{(x^2+x+1)(x^3+2x+5)}{(2x-1)^5}[/tex]
[tex]\ln y = \ln(x^2+x+1) + \ln(x^3+2x+5) - 5\ln(2x-1)[/tex]
Taking derivatives with respect to x,
[tex]y'/y = \dfrac{2x+1}{x^2+x+1} + \dfrac{3x^2+2}{x^3+2x+5} - \dfrac{10}{2x-1}[/tex]
[tex]y' = y \left( \dfrac{2x+1}{x^2+x+1} + \dfrac{3x^2+2}{x^3+2x+5} - \dfrac{10}{2x-1} \right)[/tex]
[tex]y' = \dfrac{(x^2+x+1)(x^3+2x+5)(2x-1)}{(2x-1)^6} \left( \dfrac{2x+1}{x^2+x+1} + \dfrac{3x^2+2}{x^3+2x+5} - \dfrac{10}{2x-1} \right)[/tex]
[tex]y'(2x-1)^6 =(2x+1)(x^3+2x+5)(2x-1)+(3x^2+2)(x^2+x+1)(2x-1) \\\quad -10(x^2+x+1)(x^3+2x+5)[/tex]
[tex]y'=\dfrac{-7 x^4 - 16 x^3 - 51 x^2 - 70 x - 57}{(2x-1)^6}[/tex]
