Answer
$$f'(a)=\frac{-1}{\sqrt{1-2a}}$$
Work Step by Step
$$f(x)=\sqrt{1-2x}$$
That means $$f(a)=\sqrt{1-2a}$$
The derivative of $f(x)$ at number $a$ can be found as follows: $$f'(a)=\lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a}$$
$$f'(a)=\lim\limits_{x\to a}\frac{\sqrt{1-2x}-\sqrt{1-2a}}{x-a}$$
Multiply both numerator and denominator by $\sqrt{1-2x}+\sqrt{1-2a}$, the numerator would become $$(\sqrt{1-2x}-\sqrt{1-2a})(\sqrt{1-2x}+\sqrt{1-2a})$$$$=(1-2x)-(1-2a)$$$$=2a-2x$$$$=-2(x-a)$$
Therefore, $$f'(a)=\lim\limits_{x\to a}\frac{-2(x-a)}{(x-a)(\sqrt{1-2x}+\sqrt{1-2a})}$$
$$f'(a)=\lim\limits_{x\to a}\frac{-2}{\sqrt{1-2x}+\sqrt{1-2a}}$$
$$f'(a)=\frac{-2}{\sqrt{1-2a}+\sqrt{1-2a}}$$
$$f'(a)=\frac{-2}{2\sqrt{1-2a}}$$
$$f'(a)=\frac{-1}{\sqrt{1-2a}}$$
