Answer
$2x-3$
Work Step by Step
Consider $f(x) = x^{2} - 3x +4$
We will apply definition of derivative, such as:
Thus, $f'(x) = \lim\limits_{z \to x}\dfrac{(z^{2} - 3z +4) - x^{2} + 3x -4}{(z-x)}= \lim\limits_{z \to x}\dfrac{z^{2} - 3z - x^{2} + 3x}{(z-x)}$
$\implies \lim\limits_{z \to x}\dfrac{(z+x)(z-x)-3(z-x)}{(z-x)}=\lim\limits_{z \to x}(z+x-3)=(x+x-3)$
Hence, $f'(x) = 2x-3$
