Given #f(x) = ((x^2-3x)/(2x-3))* ln(x^2 -3x)#
We can approach this by letting
#f(x) = g(x)*h(x) # # f'(x) = g(x)*h'(x) + g'(x)*h(x) " " " " " " " " " " " (1)#
#color(red)(g(x) = ((2x^2 -3x)/(2x-3))" " " " " " " " (2)#
#color(blue)(h(x) = ln(x^2 -3x) " " " " " " " " "(3)#
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Let differentiate #color(red)(g(x) = ((x^2-3x)/(2x-3))# using quotient rule
#g'(x) = ((2x-3)(2x-3)-(2(x^2-3x)))/(2x-3)^2#
#g'(x) = [(4x^2 -6x-6x+9)-(2x^2 -6x)]/(2x-3)^2#
#g'(x) = [4x^2 -12x + 9 -2x^2 +6x]/(2x-3)^2#
#color(red)(g'(x) = [2x^2-6x+9]/(2x-3)^2 " " " " " " " " " " " (4)#
Let differentiate #color(blue)(h(x) = ln(x^2 -3x)#
Note: #f(x) = ln(x) hArr f'(x) = (dx)/x#
#color(blue)(h'(x) = (2x-3)/(x^2-3x) " " " " " " " " " " (5) #
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Now we can put it together using #(1)#
#f(x) = (x^2-3x)/(2x-3) * ln(x^2 -3x)#
#f'(x) = (x^2-3x)/(2x-3) * color(blue)((2x-3)/(x^2-3x)) + color(red)((2x^2 -6x+9)/(2x-3)^2) * ln(x^2 -3x)#
#f'(x) = cancel(((x^2-3x))/(2x-3) * color(blue)((2x-3))/(x^2-3x))+ color(red)((2x^2 -6x+9)/(2x-3)^2) * ln(x^2 -3x)#
#f'(x) = 1+ [color(red)((2x^2 -6x+9)/(2x-3)^2) * ln(x^2 -3x)]#
#f'(x) = ((2x-3)^2 + (2x^2 -6x+9)* ln(x^2 -3x)]/(2x-3)^2#
