How do you find f'(x) using the limit definition given `f(x) = (x^2-1) / (2x-3)`?

Have a look:

Well it gets a bit messy.

The definition of the derivative of `y=f(x)` is

`f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h`

So with `f(x) = (x^2-1)/(2x-3)` then;

`f(x+h) = ((x+h)^2-1)/(2(x+h)-3)`
`" " = (x^2+2hx+h^2-1)/(2x+2h-3)`

And so the derivative of `y=f(x)` is given by:

`f'(t) = lim_(h rarr 0) ( ((x^2+2hx+h^2-1)/(2x+2h-3)) - ((x^2-1)/(2x-3)) ) / h`

`" " = lim_(h rarr 0) 1/h( (x^2+2hx+h^2-1)/(2x+2h-3) - (x^2-1)/(2x-3) )`

`" " = lim_(h rarr 0) ( (x^2+2hx+h^2-1)(2x-3) - (x^2-1)(2x+2h-3) ) / (h(2x+2h-3)(2x-3) )`

`" " = lim_(h rarr 0) ( (2x^3+4hx^2+2h^2x-2x-3x^2-6hx-3h^2+3) - (2x^3+2hx^2-3x^2-2x-2h+3) ) / (h(2x+2h-3)(2x-3) )`

`" " = lim_(h rarr 0) (2x^3+4hx^2+2h^2x-2x-3x^2-6hx-3h^2+3 -2x^3-2hx^2+3x^2+2x+2h+3) / (h(2x+2h-3)(2x-3) )`

`" " = lim_(h rarr 0) (2hx^2+2h^2x-6hx-3h^2+2h) / (h(2x+2h-3)(2x-3) )`

`" " = lim_(h rarr 0) (2x^2+2hx-6x-3h+2) / ((2x+2h-3)(2x-3) )`

`" " = (2x^2+0-6x-0+2) / ((2x+0-3)(2x-3) )`

`" " = (2x^2-6x-2) / ((2x-3)(2x-3) )`

`" " = (2x^2-6x-2) / (2x-3)^2`

`" " = (2(x^2-3x-1)) / (2x-3)^2`