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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.

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    No that is in the next chapter of my book. I do not know chain rules, or any of the "rules" related to derivatives. Anyways I did the problem again and got -4a/h((sqrt(1-2(a+h))) + sqrt(1-2))2011-09-11

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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 $h=0$?

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    $I$ honestly hope you prove me wrong, but in my experience, it is very difficult to do well in Calculus with weak algebra skills.2011-09-11
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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).

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    The above works as long as $\rm\:f_1 = f\:'(0)\ne 0\:.\:$ Later when you learn the derivative rules you'll see this is nothing but the special case $\rm\:n = 1/2\:$ of $\rm\:(f^{\:n})' = n\:f^{\:n-1}\:f{\:'}\:.$2011-09-12