We have to differentiate f(x) = (2x - 3)/(x^2 - 5). Use the quotient rule.

f'(x) = (2*(x^2 - 5) - (2x - 3)* 2x)/(x^2 - 5)^2

=> (2x^2 - 10 - 4x^2 + 6x)/(x^2 - 5)^2

=> (-2x^2 + 6x - 10)/(x^2 - 5)^2

The required derivative is (-2x^2 + 6x - 10)/(x^2 - 5)^2

Given that f(x)= ( 2x-3)/(x^2-5)

We need to find the first derivative f'(x)

We will use the porduct rule .

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

u= 2x-3  ==> u' = 2

v= x^2 -5 ==> v' = 2x

Then we know that:

f'(x)= ( u'v - uv')/v^2

==> f'(x)= ( 2(x^2-5) - 2x(2x-3) / (x^2-5)^2

==> f'(x)= ( 2x^2 -10 - 4x^2 + 6x)/(x^2-5)^2

==> f'(x)= ( -2x^2 + 6x -10)/(x^2-5)^2

==> f'(x)= -2(x^2 -3x +5)/ (x^2-5)^2