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First derivative

  1. Sep 7, 2008 #1
    Im trying to find the first derivative of the following equation:

    [tex]
    \frac{2}{e^x+e^{-x}}
    [/tex]

    Im trying to figure how to approach this....

    i know the derivative of ex is ex....but should I move the denominator to the numerator first?
     
    Last edited: Sep 7, 2008
  2. jcsd
  3. Sep 7, 2008 #2
    You need to use the chain rule.
     
  4. Sep 7, 2008 #3
    not to be a pest...but im a little rusty on the chain rule....how would that apply?
     
  5. Sep 7, 2008 #4

    berkeman

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    Staff: Mentor

    There are formulas for the derivitave of a product and of a quotient. Are you familiar with them? If you're not familiar with the derivative of a quotient, you can express the quotient as a product:

    [tex]\frac{A}{B} = A * B^{-1}[/tex]

    And use your usual equation for the derivative of a product.
     
  6. Sep 7, 2008 #5

    Ouabache

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    Science Advisor
    Homework Helper

    Both forms of chain rule work fine.. (product or quotient).. here's a refresher
     
    Last edited: Sep 7, 2008
  7. Sep 7, 2008 #6

    berkeman

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    Staff: Mentor

    Cool website, Ouabache!
     
  8. Sep 8, 2008 #7
    for the following problem

    [tex]

    \int {x(5x)^{-x^2}dx}

    [/tex]

    wouldnt this simplify to

    [tex]

    \int{5x^{-1}dx}

    [/tex]
     
  9. Sep 8, 2008 #8
    I don't think those two integrals are equivalent.
     
  10. Sep 8, 2008 #9
    actually I wrote the first expression wrong ....this is what the first expression should be...

    [tex]

    \int {x}{(5^{-x^2})dx}

    [/tex]
     
  11. Sep 8, 2008 #10

    Dick

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    Science Advisor
    Homework Helper

    Change the 5^(-x^2) to a power of e instead. Then use a u substitution.
     
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