Finding the derivative of a function How to find $f'(a)$ where $f(x) = \sqrt{1-2x}$ ?
I am not too sure what to do, no matter what I do I can't get the correct answer. I know it is simple algebra but I can't figure it out.
 A: As you surmise, you need to multiply by the conjugate; the problem is that you forgot to distribute the negative sign correctly, and you forgot to divide by the conjugate as well as multiply by it.
$$\begin{align*}
\lim_{h\to 0}\frac{f(a+h)- f(a)}{h} &= \lim_{h\to 0}\frac{\sqrt{1-2(a+h)}-\sqrt{1-2a}}{h}\\
&\strut\\
&=\lim_{h\to 0}\frac{\sqrt{1-2a-2h}-\sqrt{1-2a}}{h}\\
&\strut\\
&=\lim_{h\to 0}\left(\frac{\sqrt{1-2a-2h} - \sqrt{1-2a}}{h}\right)\left(\frac{\sqrt{1-2a-2h}\;+\sqrt{1-2a}}{\sqrt{1-2a-2h}\; + \sqrt{1-2a}}\right)\\
&\strut\\
&=\lim_{h\to 0}\frac{\left(\sqrt{1-2a-2h}-\sqrt{1-2a}\right)\left(\sqrt{1-2a-2h}\;+\sqrt{1-2a}\right)}{h\left(\sqrt{1-2a-2h}\;+\sqrt{1-2a}\right)}\\
&\strut\\
&= \lim_{h\to 0}\frac{(1-2a-2h) - (1-2a)}{h\left(\sqrt{1-2a-2h}\;+\sqrt{1-2a}\right)}\\
&\strut\\
&= \lim_{h\to 0}\frac{1-2a-2h-1+2a}{h\left(\sqrt{1-2a-2h}\;+\sqrt{1-2a}\right)}\\
&\strut\\
&= \lim_{h\to 0}\frac{-2h}{h\left(\sqrt{1-2a-2h}\;+\sqrt{1-2a}\right)}.
\end{align*}$$
Nothing but algebra so far. Trying to plug in $0$ for $h$ gives $\frac{0}{0}$, as expected. But there is a factor of $h$ in the numerator, and a factor of $h$ in the denominator. If we cancel them, can the resulting limit be evaluated simply by pluggin in $h=0$?
A: My answer to your prior question shows how to compute a more general derivative. Namely 
if $\rm\ f(x)\: = \ f_0 + f_1\ (x-a) +\:\cdots\:+f_n\ (x-a)^n\:$ and $\rm\: f_0 \ne 0\:$ then rationalizing the numerator below 
$$\rm \lim_{x\:\to\: a}\ \dfrac{\sqrt{f(x)}-\sqrt{f_0}}{x-a}\ = \ \lim_{x\:\to\: a}\ \dfrac{f(x)-f_0}{(x-a)\ (\sqrt{f(x)}+\sqrt{f_0})}\ =\ \lim_{x\:\to\: 0}\ \dfrac{f_1+\:\cdots\: + f_n\:(x-a)^{n-1}}{\sqrt{f(x)}+\sqrt{f_0}}\ =\ \dfrac{f_1}{2\ \sqrt{f_0}}$$
Your current problem is merely the special case $\rm\ f(x) = 1-2\:x\: =\: 1-2\:a-2\:(x-a)\:,\:$ therefore $\rm\:f_0 = 1-2\:a,\ \ f_1 = -2\:.\ $ If something about this proof is not clear then please  ask questions in the comments here or there (vs. posing more minor variants on such problems).
