1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper

    [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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar

    y = arctan(ln(x))
    tan(y) = ln(x)
    Use implicit differentiation and solve for dy/dx.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook