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Homework Help: Derivatives if ln and e

  1. Jun 14, 2007 #1
    I'm struggling taking the derivatives of anything with ln or e in it. For example the question y=xlnx/e^x,i have the solution and the first step they have is
    lny=lnx+ln(lnx)-lne^x. I understand what to do until the last part why is the e^x on top now? If anyone has the time to help me figure out this and the rest of this question that would be great because i really need to understand this.
  2. jcsd
  3. Jun 14, 2007 #2


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    Because ln(x-1) = -ln(x)

    Do you know the basic multiplication/addition properties of logarithms?
  4. Jun 14, 2007 #3
    oh right i forgot about that thanks
    what about the next part (1/y) dy/dx=1/x+1/lnx(1/x)-1 again i understand it untill the last part..i thought the derivative of e^x was just e^x..why is it 1
  5. Jun 14, 2007 #4
    never mind about that part i figured it out
  6. Jun 14, 2007 #5
    the last part i dont know how they got any of it they got xlnx/e^x[1/x+(1/xlnx)-1]
  7. Jun 15, 2007 #6


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    What "last part"? You need to ask a well-formed question (or any question for that matter) if you want an answer.
  8. Jun 15, 2007 #7


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    Write that as a function of only x and dy/dx (so no y)
  9. Jun 15, 2007 #8


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    Say you have an expression, of which derivatives is very "hard" to find using the normal way, i.e, using the Quotient Rule, the Chain Rule, Product Rule, etc... then, the next thing you should try is to use logarithms, i.e, to simplify the problem.

    Say, you want to find the derivative of:
    [tex]y = \frac{x \ln x}{e ^ x}[/tex].
    Of course, you can use a combination of the Product Rule, and Quotient Rule. But, well, it'll be messy, and you may make mistake somewhere. So it's best to take logarithm of both sides, and change all to sums (since, the Sum Rule, and Quotient Rule are far easier to use than those other rules).

    We'll have:
    [tex]\ln y = \ln \frac{x \ln x}{e ^ x}[/tex].

    Apply some properties of logarithm here:
    1. [tex]\ln \left( \frac{a}{b} \right) = \ln(a) - \ln (b)[/tex]

    2. [tex]\ln \left( ab \right) = \ln(a) + \ln (b)[/tex]

    3. [tex]\ln \left( a ^ \beta \right) = \beta \ln(a)[/tex]

    Here we go:
    [tex]\ln y = \ln \frac{x \ln x}{e ^ x} = \ln (x \ln (x)) - \ln (e ^ x)[/tex] (using (1)).

    [tex]... = \ln (x) + \ln ( \ln (x)) - \ln (e ^ x)[/tex] (using (2)).

    [tex]... = \ln (x) + \ln ( \ln (x)) - x \ln (e) = \ln (x) + \ln ( \ln (x)) - x[/tex] (Using (3)).


    So, we have:
    [tex]\ln y = \ln (x) + \ln ( \ln (x)) - x[/tex]

    If f(x) = g(x), then of course, we'll have f'(x) = g'(x), right? So, take the derivatives of both sides with respect to x, we have:

    [tex](\ln y)'_x = (\ln (x) + \ln ( \ln (x)) - x)'_x[/tex]

    [tex]\Rightarrow \frac{y'_x}{y} = (\ln (x))'_x + (\ln ( \ln (x)))'_x - (x)'_x[/tex]

    [tex]\Rightarrow \frac{y'_x}{y} = \frac{1}{x} + \frac{(\ln(x))'_x}{\ln(x)} - 1[/tex]

    [tex]\Rightarrow \frac{y'_x}{y} = \frac{1}{x} + \frac{\frac{1}{x}}{\ln(x)} - 1 = \frac{1}{x} + \frac{1}{x \ln(x)} - 1[/tex]

    Now, we have found [tex]\frac{y'_x}{y}[/tex], but well, what we need is y'x, i.e the derivatives of y with respect to x, so let's multiply both sides by y to get:

    [tex]... \Rightarrow y'_x = y \left( \frac{1}{x} + \frac{1}{x \ln(x)} - 1 \right)[/tex]

    But, what's y? Well, it's [tex]y = \frac{x \ln x}{e ^ x}[/tex], our former function.

    So, change y to x, we have:

    [tex]... \Rightarrow y'_x = \left( \frac{x \ln x}{e ^ x} \right) \times \left( \frac{1}{x} + \frac{1}{x \ln(x)} - 1 \right)[/tex]

    Yeah, we've done it. :)

    Is there anything unclear?
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