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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.2011-09-11
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    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.2011-09-11
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    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 {}.2011-09-11
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    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?2011-09-11
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    No I have no idea what a chain rule is and I have no libraries of results of anything.2011-09-11
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    @Jordan: You distributed the negative sign incorrectly, and you forgot to also *divide* by the conjugate. In short: your algebra is wrong.2011-09-11
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    Sorry I typed that out incorrectly. I did the algebra correctly on paper. I end up with something like 2a/hsquare roots. I will look at it again.2011-09-11
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    The chain rule is the rule that tells how to find the derivative of $f(g(x))$ assuming that one already knows the derivatives of $f$ and $g$ separately: $(f\circ g)'(x) = f'(g(x))g'(x)$. Perhaps you know it under another name?2011-09-11
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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

2 Answers 2

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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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    Oh I forgot you can to the h->0 thing, but how does the 2 become a one?2011-09-11
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    @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.2011-09-11
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    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.2011-09-11
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    @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?2011-09-11
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    Oh, I didn't see that for some reason. I knew they were both the same square root but I just didn't see that for some reason. Probably for the same reason I will fail this test.2011-09-11
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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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    So basically any problem involving squareroots for f(x) is going to be n/2sqrtx?2011-09-12
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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