Answer
$f'(x) = 2x - 4$
Work Step by Step
$f(x) = x^{2} -4x + 5$
$f'(x) = \lim\limits_{h \to 0} \frac{(x+h)^{2} - 4(x+h) + 5 - (x^{2} - 4x+5)}{h}$
$f'(x) = \lim\limits_{h \to 0} \frac{x^{2} + 2xh + h^{2} -4x-4h + 5 - x^{2} +4x-5}{h}$
$f'(x) = \lim\limits_{h \to 0} \frac{2xh + h^{2} -4h}{h}$
$f'(x) = \lim\limits_{h \to 0} \frac{h(2x + h -4)}{h}$
$f'(x) = \lim\limits_{h \to 0} (2x + h -4)$
$f'(x) = \lim\limits_{h \to 0} (2x + (0) -4)$
$f'(x) = 2x - 4$
