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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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Can you show how you get an incorrect answer? Then it would be easier to pinpoint which mistake you need to correct. – Henning Makholm Sep 11 '11 at 22:47
Sure. I know I have to use f(a+h) -f(a) / h so that gives me sqrt(1-2(a+h) - sqrt (1-2(a)) / h I am guessing that I use a conjugate so that rules in 1-2a-2h - 1 - 2a / h which gives me the wrong answer. – user138246 Sep 11 '11 at 22:47
The body of your question doesn't contain a question, and the title is an incomprehensible fragment. One can only guess that you're trying to differentiate this function? By the way, you can get the square root to extend over its argument by enclosing the latter in curly brackets {}. – joriki Sep 11 '11 at 22:49
By the time you're asked to differentiate things like this, you should have a library of results you can use to differentiate symbolically without going back to the original definition each time. Does "chain rule" ring any bells? – Henning Makholm Sep 11 '11 at 22:50
No I have no idea what a chain rule is and I have no libraries of results of anything. – user138246 Sep 11 '11 at 22:50
up vote 5 down vote accepted

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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Oh I forgot you can to the h->0 thing, but how does the 2 become a one? – user138246 Sep 11 '11 at 23:01
@Jordan: You forgot to take the limit? That's a fairly important part of finding the derivative... Once you cancel and do the evaluation, you should end up with a factor of $2$ in both the numerator (the one you see already) and the denominator; they will then cancel. – Arturo Magidin Sep 11 '11 at 23:03
I don't think I am following. The next step I see is cancelling out the h and then I am left with two square roots. – user138246 Sep 11 '11 at 23:04
@Jordan: And then you plug in $h=0$ (that is, you take the limit). You are left with$$\frac{-2}{\sqrt{1-2a-0}+\sqrt{1-2a}} = \frac{-2}{\sqrt{1-2a}+\sqrt{1-2a}} = \frac{-2}{2\sqrt{1-2a}}.$$No? – Arturo Magidin Sep 11 '11 at 23:05
I honestly hope you prove me wrong, but in my experience, it is very difficult to do well in Calculus with weak algebra skills. – Arturo Magidin Sep 11 '11 at 23:20

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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So basically any problem involving squareroots for f(x) is going to be n/2sqrtx? – user138246 Sep 12 '11 at 1:04
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 $\rm\:(f^{\:n})' = n\:f^{\:n-1}\:f{\:'}\:.$ – Bill Dubuque Sep 12 '11 at 1:13

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