Question

Please provide an explanation and steps

Answer 1

Haven't done one of these in forever.  We avoid the quotient rule with an implicit logarithmic differentiation.

$y=\dfrac{(x^2+x+1)(x^3+2x+5)}{(2x-1)^5}$

$\ln y = \ln(x^2+x+1) + \ln(x^3+2x+5) - 5\ln(2x-1)$

Taking derivatives with respect to x,

$y'/y = \dfrac{2x+1}{x^2+x+1} + \dfrac{3x^2+2}{x^3+2x+5} - \dfrac{10}{2x-1}$

$y' = y \left( \dfrac{2x+1}{x^2+x+1} + \dfrac{3x^2+2}{x^3+2x+5} - \dfrac{10}{2x-1} \right)$

$y' = \dfrac{(x^2+x+1)(x^3+2x+5)(2x-1)}{(2x-1)^6} \left( \dfrac{2x+1}{x^2+x+1} + \dfrac{3x^2+2}{x^3+2x+5} - \dfrac{10}{2x-1} \right)$

$y'(2x-1)^6 =(2x+1)(x^3+2x+5)(2x-1)+(3x^2+2)(x^2+x+1)(2x-1) \\\quad -10(x^2+x+1)(x^3+2x+5)$

$y'=\dfrac{-7 x^4 - 16 x^3 - 51 x^2 - 70 x - 57}{(2x-1)^6}$