Answer
$f^{'}(x) = \frac{2x^{2}-6x +2 }{(2x-3)^{2}}$
$\left\{x \in R| x \not=\frac{3}{2}\right\}$
Work Step by Step
The definition of the derivation of a function is : $f^{'}(x) = \lim\limits_{h \to 0}\frac{f(x+h) -f(x) }{h}$ ---------------- Apply the definition of the derivation of a function to $f(x) = \frac{x^{2} ~~-~~ 1}{2x~~ - ~~3}$
$f^{'}(x) = \lim\limits_{h \to 0}\frac{\frac{~~(x+h) ^{2} -~~1}{~~~2(x~~+~~h)~~ -~~3}-\frac{x^{2}~~-~~1}{~2x~~-~~3}}{h}$ $f^{'}(x) = \lim\limits_{h \to 0}\frac{(2x-3)[(x+h)^{2}-1]~~-~~(x^{2}~-~1)[2(x~+~h)~-~3]}{(~2x~-~3~)[~(2(x~+~h)~-~3~]h}$ $f^{'}(x) = \lim\limits_{h \to 0}\frac{(2x^{3}~+~4x^{2}h~+~2xh^{2}~-~2x~-~3x^{2}-~6xh-~3h^{2}+~3)~-~(2x^{3}~+~2x^{2}h~-~3x^{2}~-~2x~-~2h~+~3~)}{(~2x~-3~)~[~2~(~x~+~h~)~-~3~]~h}$ $f^{'}(x) = \lim\limits_{h \to 0}\frac{2x^{2}h~ +~ 2xh^{2}~-~6xh~-~3h^{2}~+~2h}{(~2x~-~3~)[~2(~x~+~h)-3~]h}$
$f^{'}(x) = \lim\limits_{h \to 0}\frac{-2x^{2}~ +~ 2xh~-~6x~-~3h~+~2}{(~2x~-~3~)[~2(~x~+~h)-3~]}$
$f^{'}(x) =~\frac{2x^{2}~-~6x~+~2}{(~2x~-~3~)^{2}}$
The domain of the function and its first derivative is:
$\left\{x \in R| x \not=\frac{3}{2}\right\}$
