Find f'(x) if f(x) = (x-2)/(2x-3)

We have f(x) = ( x - 2)/ ( 2x -
3)

=> f(x) = (x - 2) * ( 2x -
3)^-1

We use the product rule which states that if h(x) =
f(x)*g(x)

=> h'(x) = f(x)*g'(x) +
f'(x)*g(x)

f(x) = (x - 2) * ( 2x -
3)^-1

=> f'(x) = [(x - 2)]' * ( 2x - 3)^-1 + (x - 2)
* [( 2x - 3)^-1]'

=> f'(x) = 1* ( 2x - 3)^-1 - (x -
2) * 2 * (2x - 3)^-2

=> f'(x) = [(2x - 3) - 2x + 4]/
(2x - 3)^2

=> f'(x) = 1/ (2x -
3)^2

Therefore for f(x) = (x - 2)/(2x - 3),
f'(x) = 1/ (2x - 3)^2