y=ln(x^2-3x+2)

differentiate y=ln(x^2-3x+2)
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1 Answer

y = ln (x^2 +-3x + 2)

let U = x^2 - 3x + 2

du/dx = 2x - 3

y = lnU

dy/du = 1/u

dy/du = 1/(x^2 - 3x + 2)

from the chain rule 

dy/dx = dy/du * du/dx

dy/dx = {1/(x^2 - 3x + 2)}*{2x - 3}

=(2x - 3)/(x^2- 3x + 2)

dy/dx = (2x - 3)/(x^2- 3x + 2)

by Level 3 User (4.0k points)

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