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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. – Henning Makholm Oct 13 '11 at 21:08
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. – user138246 Oct 13 '11 at 21:09
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. – Arturo Magidin Oct 13 '11 at 21:12
@Jordan: I changed the title because "logarithmic differentiation" actually has a specific meaning, and this isn't it... – Arturo Magidin Oct 13 '11 at 21:15
@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. – Arturo Magidin Oct 13 '11 at 21:23
up vote 5 down vote accepted

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. – user138246 Oct 13 '11 at 21:20


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

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I still get the wrong answer. I am getting $f(x)\prime ln(x^-1) = ln-1x^-2$ – user138246 Oct 13 '11 at 21:16
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. – Arturo Magidin Oct 13 '11 at 21:21
I do not see what you mean? Do you know some logarithmic rules? – AD. Oct 13 '11 at 21:21
@Arturo Magidin: :) – AD. Oct 13 '11 at 21:22
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. – user138246 Oct 13 '11 at 21:24

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 $-1/2x$ – user138246 Oct 13 '11 at 21:46

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