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Find the derivative of: 1/1 + e^-x

  1. Apr 8, 2008 #1
    Here's my question:
    Find the derivative of:
    1/1 + e^-x

    Here is my attemp, since 1/x=lnx and -x=-1, => ln(1-e^-x)
    I don't know if this is right, I would appreciate some guidance.
  2. jcsd
  3. Apr 8, 2008 #2
    well i am assuming you meant the following


    you need to apply the quotient rule here, do you know the quotient rule?

    here it is in case you have forgotten it:

  4. Apr 8, 2008 #3
    Or you can write the function as: [tex](1+e^{-x})^{-1}[/tex]

    And then apply the power rule: [tex]\frac{d}{dx} u^n=nu^{n-1} \frac{du}{dx}[/tex]
  5. Apr 8, 2008 #4
    Ok, using the quotient rule:

    So we have:

    http://img406.imageshack.us/img406/5082/sdfth1nz7.gif [Broken]

    Is that right?

    But using the other method I get:
    So: (-e^-x) . -1 . (1+e^-x)^-2
    inner function is: -e^-x

    Which answer do you think is right?

    Thank you.
    Last edited by a moderator: May 3, 2017
  6. Apr 8, 2008 #5
    what on earth is this?
  7. Apr 8, 2008 #6


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    [tex]\frac{0\times \left(e^{-x}+1\right)+e^{-x}\times

    roam, next time please use TeXForm and put it between [tex], [/ tex] tags (without the space). And if you have Mathematica anyway, you can use it to check the answer:
    FullSimplify[....youranswer... == D[ function, x ]]
  8. Apr 9, 2008 #7
    Thank you Compuchip, I understand it now and by simplifying it we get:
    e^-x / (1+e^-x)^2

    But I have another question:
    http://img170.imageshack.us/img170/1766/xcvdxf1fs3.gif [Broken]
    Let y = tan-1x
    Then, Let x = tany
    dx/dy = sec2y = 1+tan2y [Trigonometric identity]
    Therefore dy/dx = 1/(1+tan2y) = 1/(1+x2)

    Am I on the right track?

    Thank you. :smile: :smile:
    Last edited by a moderator: May 3, 2017
  9. Apr 10, 2008 #8


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    Not really. First of all, [itex]\tan^{-1}(\cdots)[/itex] can mean either [itex]1/\tan(\cdots)[/itex], so evaluate the tan function and then take its reciprocal. Or it can mean [itex]\arctan(\cdots)[/itex], which is the function such that if you take tan(arctan(x)) you get x back.

    Since the derivative of the second one is hard do find by the methods you have been using so far, I'll just assume you meant to write
    [tex]y = \frac{1}{\tan(\ln x)}[/tex]
    For this, I'd suggest using the chain rule:
    [tex]u(x) = \tan(\ln x) \qquad\Rightarrow\qquad y = u^{-1}, \frac{dy}{dx} = \left(\frac{dy}{du} \right) \left( \frac{du}{dx} \right).[/tex]
    Last edited: Apr 10, 2008
  10. Apr 10, 2008 #9


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    y = arctan(ln(x))
    tan(y) = ln(x)
    Use implicit differentiation and solve for dy/dx.
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