use logarithmic differentiation to find the derivative of y=(2x-1)/(3x-2), x>1

Question

use logarithmic differentiation to find the derivative of y=(2x-1)/(3x-2), x>1

Answer 1

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

Take natural log of both sides

ln(y) = ln(2x-1) - ln(3x-2)

Take the derivative of both sides

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

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

dy/dx = -1/(3x-2)²