I was using log rules and was able to solve to the point y'=y(1/2x + 2x-1 +2/3(x+1) how did my professor simplify it to the below answer?

Question

1/6 x ^-1/2 e ^{x^2-x} (x+1)^-1/3 (12x³+6x²+x+3)

Answer 1

y = x^1/2e^{x^2 - x}(x + 1)^2/3 Use logarithmic differentiation to find y':

lny = (1/2)lnx + x² - x + (2/3)ln(x + 1)

y' / y = 1 / (2x) + 2x - 1 + 2 / (3(x+1))

y' = y ( [3(x + 1) + 6x(x + 1)(2x - 1) + 4x] / [6x(x + 1)]

y' = y [ 12x³ + 6x² + x + 3 ] / [6x(x + 1)]

So, y' = x^1/2e^{x^2 - x}(x + 1)^2/3 [ 12x³ + 6x² + x + 3 ] / [6x(x + 1)]

Simplify to obtain y' = (1/6)x^-1/2e^{x^2 - x} (x + 1)^-1/3 [ 12x³ + 6x² + x + 3 ]