Answer
$f'(a) = -\dfrac{1}{\sqrt {1-2a}}$
Work Step by Step
Given $f(x) = \sqrt{1-2x}$
$f'(a) = \lim\limits_{x \to a} \dfrac{f(x) - f(a)}{x-a} = \lim\limits_{x \to a} \dfrac{\sqrt{1-2x}- \sqrt{1-2a}}{x-a} = \lim\limits_{x \to a}\bigg[ \dfrac{\sqrt{1-2x}- \sqrt{1-2a}}{x-a} \times \dfrac{\sqrt{1-2x} + \sqrt{1-2a}}{\sqrt{1-2x} + \sqrt{1-2a}}\bigg] = \lim\limits_{x \to a} \dfrac{(\sqrt{1-2x}- \sqrt{1-2a})(\sqrt{1-2x} + \sqrt{1-2a})}{(x-a)(\sqrt{1-2x} + \sqrt{1-2a})} = \lim\limits_{x \to a} \dfrac{(\sqrt{1-2x})^2 - (\sqrt{1-2a})^2}{(x-a)(\sqrt{1-2x} + \sqrt{1-2a})} = \lim\limits_{x \to a} \dfrac{(1-2x) - (1-2a)}{(x-a)(\sqrt{1-2x} + \sqrt{1-2a})} = \lim\limits_{x \to a} \dfrac{1-2x -1+2a}{(x-a)(\sqrt{1-2x} + \sqrt{1-2a})} = \lim\limits_{x \to a} \dfrac{-2x + 2a}{(x-a)(\sqrt{1-2x} + \sqrt{1-2a})} = \lim\limits_{x \to a} \dfrac{-2(x - a)}{(x-a)(\sqrt{1-2x} + \sqrt{1-2a})} = \lim\limits_{x \to a} \dfrac{-2}{\sqrt{1-2x} + \sqrt{1-2a}} = -\dfrac{2}{\sqrt{1-2a} + \sqrt{1-2a}} = -\dfrac{2}{\sqrt{1-2a} + \sqrt{1-2a}} = -\dfrac{2}{2\sqrt{1-2a}} = -\dfrac{1}{\sqrt{1-2a}} \longrightarrow f'(a) = -\dfrac{1}{\sqrt {1-2a}}$
