Calculus: Early Transcendentals 8th Edition

$y'=-\dfrac{(x^{2}+1)^{3}(x^{2}+56x+9)}{(2x+1)^{4}(3x-1)^{6}}$
$y=\dfrac{(x^{2}+1)^{4}}{(2x+1)^{3}(3x-1)^{5}}$ Start the differentiation process by using the quotient rule: $y'=...$ $...=\dfrac{(2x+1)^{3}(3x-1)^{5}[(x^{2}+1)^{4}]'-(x^{2}+1) ^{4}[(2x+1)^{3}(3x-1)^{5}]'}{[(2x+1)^{3}(3x-1)^{5}]^{2}}=...$ $...=\dfrac{(2x+1)^{3}(3x-1)^{5}[(x^{2}+1)^{4}]'-(x^{2}+1) ^{4}[(2x+1)^{3}(3x-1)^{5}]'}{(2x+1)^{6}(3x-1)^{10}}=...$ Use the chain rule to evaluate $[(x^{2}+1)^{4}]'$ and the product rule and chain rule to evaluate $[(2x+1)^{3}(3x-1)^{5}]'$: $...=\dfrac{(2x+1)^{3}(3x-1)^{5}[4(x^{2}+1)^{3}(2x)]-(x^{2}+1)^{4}[(2x+1)^{3}(5)(3x-1)^{4}(3)+(3x-1)^{5}(3)(2x+1)^{2}(2)]}{(2x+1)^{6}(3x-1)^{10}}=...$ $...=\dfrac{8x(2x+1)^{3}(3x-1)^{5}(x^{2}+1)^{3}-(x^{2}+1)^{4}[15(2x+1)^{3}(3x-1)^{4}+6(3x-1)^{5}(2x+1)^{2}]}{(2x+1)^{6}(3x-1)^{10}}=...$ $...=\dfrac{8x(2x+1)^{3}(3x-1)^{5}(x^{2}+1)^{3}-15(x^{2}+1)^{4}(2x+1)^{3}(3x-1)^{4}-6(x^{2}+1)^{4}(3x-1)^{5}(2x+1)^{2}}{(2x+1)^{6}(3x-1)^{10}}=...$ Take out common factor $(2x+1)^{2}(3x-1)^{4}(x^{2}+1)^{3}$ from the numerator and simplify: $...=\dfrac{(2x+1)^{2}(3x-1)^{4}(x^{2}+1)^{3}[8x(2x+1)(3x-1)-15(x^{2}+1)(2x+1)-6(x^{2}+1)(3x-1)]}{(2x+1)^{6}(3x-1)^{10}}=...$ $...=\dfrac{(x^{2}+1)^{3}(48x^{3}+8x^{2}-8x-30x^{3}-15x^{2}-30x-15-18x^{3}+6x^{2}-18x+6)}{(2x+1)^{4}(3x-1)^{6}}=...$ $...=\dfrac{(x^{2}+1)^{3}(-x^{2}-56x-9)}{(2x+1)^{4}(3x-1)^{6}}=...$ $...=-\dfrac{(x^{2}+1)^{3}(x^{2}+56x+9)}{(2x+1)^{4}(3x-1)^{6}}$