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Differentiation of e^(-x) / ln(x)

  1. Aug 6, 2015 #1
    Hi there
    Getting stuck on this equation.
    e^(-x) / ln(x)

    Solving it by quotient rule, however answer has extra x in numerator.
    using the dy/dx = (v(du/dx)-u(dv/dx))/v^2

    dy/dx = (ln(x)*-e^(-x) - (-e^(-x)*1/x)/ln(x)^2

    Answer = (-e^(-x) (x ln(x)-1))/x(ln(x))^2

    Would appreciate help with breaking this down. Cheers
     
  2. jcsd
  3. Aug 7, 2015 #2

    fzero

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    Your result has a factor
    $$ \ln x - \frac{1}{x} = \frac{ x\ln x -1}{x}$$
    so it is equivalent to the form given as the answer.
     
  4. Aug 7, 2015 #3
    Thank you for the reply. Sorry, not sure that I understand your response. Can you please break it down, to show where the "x" value is coming from.

    Cheers.
     
  5. Aug 7, 2015 #4

    fzero

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    You wrote dy/dx = (ln(x)*-e^(-x) - (-e^(-x)*1/x)/ln(x)^2, so if I format this we have
    $$ \frac{ \ln x (-e^{-x}) - (-e^{-x}) \frac{1}{x} }{ (\ln x)^2} = - \frac{e^{-x}}{ (\ln x)^2} \left( \ln x - \frac{1}{x} \right).$$
    Now we can write
    $$ \ln x - \frac{1}{x} = \frac{x}{x} \ln x - \frac{1}{x} = \frac{ x \ln x - 1}{x}.$$
    When we put this into the previous expression, we get
    $$ - \frac{e^{-x}}{ (\ln x)^2} \frac{ x \ln x - 1}{x} = - e^{-x}\frac{ x \ln x - 1}{x(\ln x)^2},$$
    which is the same as the answer that you wrote as (-e^(-x) (x ln(x)-1))/x(ln(x))^2.
     
  6. Aug 7, 2015 #5
    Many thanks for the response. Thank you. Now I get it.
    Cheers
     
  7. Aug 7, 2015 #6

    I made a mistake...should be
    You wrote dy/dx = (ln(x)*-e^(-x) - (e^(-x)*1/x)/ln(x)^2

    Cheers Petra
     
  8. Aug 7, 2015 #7

    fzero

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    Sorry I should have caught that. From the start we have
    $$ \begin{split} \frac{d}{dx} \frac{e^{-x}}{\ln x} &= \frac{1}{\ln x} \frac{d}{dx} e^{-x}+e^{-x} \frac{d}{dx} \frac{1}{\ln x} \\
    & = - \frac{e^{-x}}{\ln x} - e^{-x} \frac{1}{x (\ln x)^2} \\
    & = - e^{-x} \frac{1}{x (\ln x)^2} ( x\ln x +1). \end{split}$$
    Are you sure there isn't a + sign in the answer that was given?
     
  9. Aug 10, 2015 #8
    No, I've double checked the answer. It has a "-" which I don't think is correct.

    Cheers Petra
     
  10. Aug 12, 2015 #9

    Mark44

    Staff: Mentor

    Minor point -- the above is not an equation, with the main clue being that there is no =. An equation indicates that two quantities have the same value; i.e., are equal.
     
  11. Aug 12, 2015 #10
    My mistake, used the wrong forum category and terminology.
     
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