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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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    @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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    @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