Y=[(4x+1)√(x²-3x)]/x²
ln y=ln{[(4x+1)√(x²-3x)]/x² }
ln y=ln(4x+1)+(1/2)ln(x²-3x)-2lnx
(1/ y)·y' = [1/(4x+1)]·4+(1/2)·[1/(x²-3x)]·(2x-3)-2/x
y'=y·{ 4/(4x+1)+(2x-3)/[2(x²-3x)]-2/x}
y'={[(4x+1)√(x²-3x)]/x²} ·{ 4/(4x+1)+(2x-3)/[2(x²-3x)]-2/x}
или
y=[(4x+1)√(x²-3x)]/x²
y'= {[(4x+1)√(x²-3x)]'x²-(x²)'· [(4x+1)√(x²-3x)]}/(x² )²