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

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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**

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

To find the derivative, we will use the quotient rule.

Let f(x) = u/v such that:

u= x-2 ==> u' = 1

v= (2x-3) ==> v' = 2

Then we know that:

f'(x) = [ u'*v - u*v']/ v^2

= [1*(2x-3) - (x-2)*2 ] / (2x-3)^2

= (2x-3 -2x+4) / (2x-3)^2

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