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I am attempting to review this material and differentiate $f(x)=\ln(1/x)$

I know that $(\ln x)'= 1/x\ $ but this just seems to complicate the problem and I don't think it will assisst me in solving it. I think what I am suppose to do is differentiate in a different way but I don't know how. I went back through the chapter and they use some incredibly complex piece-wise defined functions for the definition and basically just tell me to not worry about it and just memorize $1/x$. How am I supposed to approach this problem?

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    Hint: chain rule.2011-10-13
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    I tried that but I must be doing something incorrectly as I do not get the answer the book does. I tried everything I can think of.2011-10-13
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    Everything you can think of ought to be quite a lot. Care to show some of it?2011-10-13
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    No, you don't know that $\ln x = \frac{1}{x}$ (and if you *do* "know it", then you know something that is false). What you *may* know is that $(\ln x)'$ (the **derivative** of $\ln x$) is equal to $\frac{1}{x}$. Again: this is not a trivial error, it's a reflection of not keeping the distinction between a function and its derivative clear.2011-10-13
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    Keeping secret your wrong answer is not the way to get help here!2011-10-13
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    @AD.: That correction may not have been a "trivial" one; the OP has a long track record of such errors, and they tend to reverberate to create problems down the line.2011-10-13
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    @Jordan: I changed the title because "logarithmic differentiation" actually has a specific meaning, and this isn't it...2011-10-13
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    @Arturo Magidin: I just thought it was an obvious typo.2011-10-13
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    @AD.: Yes; but if you see, for example, the OPs comment below, he often writes `function` = `derivative of the function`, or similar "stream-of-consciousness-chains-of-equalities". So it was unlikely to be a typo, and it's the kind of thing that just helps trip him up later.2011-10-13

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Even simpler hint: $$\ln(a/b) = \ln(a)-\ln(b),$$ so $\ln(1/x) = $insert answer here.

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    Ugh, I can see the answer without doing the work. I forgot about the log rules.2011-10-13
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Hint:

You might start with $1/x=x^{-1}$.

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    I still get the wrong answer. I am getting $f(x)\prime ln(x^-1) = ln-1x^-2$2011-10-13
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    You are: (i) misapplying the Chain Rule; and (ii) once again writing that the function equals its derivative. You did $(f(g(x)))' = f'(g'(x))$ with $f(u) = \ln(u)$ and $g(x)=x^{-1}$. Remember, the Chain Rule says$$(f(g(x)))' = f'(g(x))g'(x).$$ Use that with $f(u)=\ln u$ and $g(x) = x^{-1}$; and *please* stop writing an equal sign between a function and its derivative.2011-10-13
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    I do not see what you mean? Do you know some logarithmic rules?2011-10-13
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    @Arturo Magidin: :)2011-10-13
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    I forgot all the log rules. I just don't understand how I can use the chain rule here actually. I have ln, ln means nothing without x right? It is like sin, sin is nothing unless I have something after it.2011-10-13
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    @Jordan: You have the rule $\ln x^a=a\ln x$ -- use this for $a=-1$ before looking at the derivative...2011-10-13
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    I don't know about $\ln x^a=a\ln x$2011-10-13
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    @Jordan: This is one of the main things about the logarithm. I suggest you study the exponential function as well as the logarithm before going into derivatives of these functions.2011-10-13
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    @Jordan: That's a basic log rule. You may not remember it, but I'll bet dollars to donuts that it was one of the "log rules" you were taught. As to how to use a Chain Rule, you have $f(u)=\ln(u)$ and $g(x) = x^{-1}$. Then $f'(u) = \frac{1}{u}$, $g'(x) = -x^{-2}$, so $$(f(g(x)))' = f'(g(x))g'(x) = \frac{1}{g(x)}g'(x) = \frac{1}{x^{-1}}(-x^{-2}) = (x^1)(-x^{-2}) = -x^{-1} = -\frac{1}{x}.$$2011-10-13
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    @Jordan: Note that you don't have to use the chain rule. My hint was for you to see that $$f(x)=\ln(1/x)=\ln(x^{-1})=(-1)\cdot\ln x=-\ln x$$ and then you get $$f'(x)=(-\ln x)'=-(\ln x)'=-1/x.$$2011-10-13
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    I don't understand how you can use the chain rule on something like lnx since it isn;t a composition it is just a single term.2011-10-13
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    @Jordan: (i) The suggestion to use the Chain Rule is not on $\ln(x)$, it is on $\ln(1/x)$, which *is* a composition: it's the function $f(u)=\ln u$ composed with the function $g(x)=1/x$. Just because you can *also* write this function in a different way in which it does not seem to equal a composition does not mean it is not a composition when expressed this way. (ii) You can always use the Chain Rule on a function like $f(x)=\ln(x)$: it's the composition of $f(u)=\ln(x)$ and $g(x)=x$. It's just that in that case, the Chain Rule is a waste of effort: $g'(x)=1$, so you get $f'(g(x))g'(x)=f'(x)$2011-10-14
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Another hint: Write $g(x) = 1/x$. How would you differentiate $\ln(g(x))$?

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    I don't quite understand the notation but I would just have 0/1 for the derivative. Or maybe I need to use the quotient rule which would give me $1/x^2$ or I could just use the quick method which would give me $-1x^-2$ which might be something like $-1/2x$2011-10-13