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Homework Help: Finding derivative of ln(x)

  1. May 16, 2010 #1
    1. The problem statement, all variables and given/known data

    I'm trying to prove that d ln(x) / dx = 1/x

    This isn't a homework problem of mine for any class. I'm just doing it for fun, so if I'm faced with something I'm not sure of, I apologize. I've only made it through Calculus 2

    3. The attempt at a solution

    Difference quotient

    ln(x+h)-ln(x) / h

    ln([x+h]/x) / h

    ln(1+[h/x]) * 1/h

    u = h/x So limit h-->0 becomes limit u--> 0

    ln(1+u) * 1/ux = 1/x * ln[(1+u)^(1/u)]

    Here's where I stopped. A friend of mine told me the ln[(1+u)^(1/u)] as u approaches 0 = ln(e) which makes sense, and I believe he said it was a known identity. Can anyone prove this fact to me?
  2. jcsd
  3. May 16, 2010 #2


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    you can do it using L'hop but that implies you already know the derivative of ln(x)... though perhaps you could use the argument leading to L'Hops or just the definintion of e

    though its not strictly 1st principles how about starting from the definition:
    [tex] e^{ln(x)} = x [/tex]
    then differentiating using chain rule & derivative of e^x?
    Last edited: May 16, 2010
  4. May 16, 2010 #3


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    A different proof is constructed using the fact that:
    [tex]y=ln \left( x \right)[/tex] can be rearranged as [tex]e^{y}=x[/tex]

    The rest of this derivative derivation is left as an exercise to the reader! (But I suggest using implicit differentiation)
  5. May 16, 2010 #4

    Char. Limit

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    Simple... define v as 1/u, and put that in. Then you get ln[(1+(1/v))^v] as v goes to infinity. e is defined as the inner number.
  6. May 16, 2010 #5
    I can prove that [itex](1 + x)^{\frac{1}{x}}[/itex] has a limit when [itex]x \rightarrow 0^{+}[/itex]. Then, you tell me the definition of [itex]e[/itex] that you have accepted, and I will show you that limit is the number [itex]e[/itex].
  7. May 17, 2010 #6
    Thanks everyone, it makes a lot of sense using simpler methods on easier examples such as e^y=x or e^(lnx)=x

    lim x-> 0 (1+x)^(1/x) and lim x-> infinity (1+[1/x])^x both being equal to e is still a bit mysterious to me. What does this have to do with? Sequences and series possibly?
  8. May 17, 2010 #7


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    try dickfore's suggestion....

    you have to start with a definition of e. whats yours?

    For example, say you start with:
    [tex] e = \lim_{n \to \infty} (1 + \frac{1}{n})^n[/tex]
    (as dickfore suggests, you can show this is increasing & bounded so must have a limit... we'll call it e)

    then multiplying that out, using binomial theorem, you get
    [tex] e
    = \lim_{n \to \infty} \sum_{j=0}^{n} \begin{pmatrix} n \\ j \end{pmatrix} \frac{1}{n^j}
    = \lim_{n \to \infty} \sum_{j=0}^{n} \frac{n!}{j!(n-j)!}\frac{1}{n^j}
    = \lim_{n \to \infty} \sum_{j=0}^{n} \frac{1}{j!}\frac{\sum_{k=0}^{j} n^{j-k}}{n^j}
    = \sum_{j=0}^{\infty} \frac{1}{j!} [/tex]
    which is the more familiar expansion of [itex] e^1 [/itex]

    similarly once you label one limit e, you can show others are equivalent
  9. May 17, 2010 #8


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    I will also point out that many modern texts define ln(x) to be
    [tex]\int_1^x \frac{1}{t}dt[/tex]
    for x> 0, and then derive all the properties of ln(x) as well as the fact that it is the inverse to [math]e^x[/math] (where e is the number such that ln(e)= 1). From that it follows trivially that d(ln(x))/dx= 1/x.

    If, instead, you have defined ln(x) to be the inverse function to [itex]e^x[/itex], then you need to first show that [itex]d e^x/dx= e^x[/itex] and use the fact that, since y= ln(x) implies that [itex]x= e^y[/itex],

    [tex]\frac{d ln(x)}{dx}= \frac{dy}{dx}= \frac{1}{\frac{dx}{dy}}= \frac{1}{\frac{d e^y}{dy}}= \frac{1}{e^y}= \frac{1}{x}[/tex]
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