Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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
1  
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
1  
@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
1  
@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

3 Answers 3

up vote 4 down vote accepted

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

share|improve this answer
    
Ugh, I can see the answer without doing the work. I forgot about the log rules. –  user138246 Oct 13 '11 at 21:20

Hint:

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

share|improve this answer
    
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 $\ln(g(x))$?

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.