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Differentiating ln x from first principles

  1. Jul 7, 2003 #1
    Hi all.

    For any of you who have done differential calculus, I need a little help with a problem involving natural logarithms.

    The question asks to differentiate y = ln x from first principles . It says "use the definition of the Euler number, namely e = lim(n->inf.) (1+1/n)^n.".
    First principles means f'(x) = lim(h->0) [f(x+h) - f(x)] / h (this is the first thing we learned in calculus).

    I so far managed two different methods:
    Method 1. y = ln x
    therefore e^y = x
    dx/dy = e^y.
    Since dx/dy * dy/dx = 1
    1/(dx/dy) = dy/dx.
    = 1/e^y
    = 1/x.

    Method 2. y = ln x
    f(x) = ln x
    f(x+h) = ln (x+h)
    f'(x) = lim(h->0) [ln (x+h) - ln x] / h
    = lim(h->0) ln (x+h/x) / h
    = lim(h->0) 1/h * ln(1+h/x)
    Since lim(h->0) ln(1+h/x) -> h/x where h != 0,
    f'(x) = 1/h * h/x
    = 1/x

    Both of these methods work and are valid, although I didn't bring the definition of the Euler number into it. I personally have no idea how to do this. Could anyone here who has done a bit of math before please help me with this?

    Thanks.
     
  2. jcsd
  3. Jul 7, 2003 #2

    HallsofIvy

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    Would you mind telling me how you arrive at
    ?
    What happened to the first x?

    Not to mention:
    Surely you know that lim(h->0) does NOT depend on h!


    Probably what your instructor intends is to use your first calculation: if y= ln(x) then x= e^y . HOWEVER, you cannot assume that the derivative of e^x is e^x: that's where "first principles" comes in.

    If y(x)= e^x, then y(x+ h)= e^(x+h) so (y(x+h)- y(x))/h=
    (e^(x+h)-e^x)/h= (e^x e^h- e^x)/h= (e^x)((e^h-1)/h).

    The derivative of e^x is (lim(h->0)(e^h-1)/h) e^x.

    You need to show "from first principles" that

    lim(h->0) (e^h-1)/h = 1.
     
  4. Jul 7, 2003 #3
    The first "x" wasn't lost- remember the logarithm laws:
    ln a - ln b = ln(a/b) so my expression ln(x+h) - ln(x) became ln(x+h/x).

    As for differentiating e^x, that is easy.
    f(x) = e^x
    f(x+h) = e^(x+h)
    f'(x) = lim(h->0) e^(x+h) - e^(x) / h
    = lim(h->0) e^x(e^h - 1) / h
    = e^x lim(h->0) e^h - 1 /h
    The definition of the Euler number is lim(n->inf.) (1+1/n)^n which can become lim(h->0) (1+h)^1/h.
    therefore
    f'(x) = e^x lim(h->0) (1+h)^1/h^h - 1 / h
    = e^x lim(h->0) 1 + h - 1 /h
    = e^x lim(h->0)h/h
    = e^x

    Although that doesn't really come into it. A previous question in the excercise asked that, and I think that there is probably some other method which needs to be used. Is anyone here familiar with any other methods?
     
  5. Jul 7, 2003 #4

    HallsofIvy

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    Oh, I see now: when you wrote ln(x+h/x) you MEANT ln((x+h)/x)
    (although when you wrote ln(1+ h/x) you DIDN'T mean ln((1+h)/x).)
     
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