Question

What is the derivative of `f(x)=((x^2-3x)/(2x-3))*ln(x^2-3x)`?

Answer 1

`f'(x) = ((2x-3)^2 + (2x^2 -6x+9)* ln(x^2 -3x)]/(2x-3)^2`

Explanation:

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`