Results 1 to 5 of 5

Math Help - exp^x-exp^-x / exp^x+exp^-x=exp^2*x-1/exp^2*x+1 --> hyperbolic tangent

  1. #1
    Junior Member
    Joined
    Feb 2010
    Posts
    50

    exp^x-exp^-x / exp^x+exp^-x=exp^2*x-1/exp^2*x+1 --> hyperbolic tangent

    I dont have a clue how people come up to the second part of the equation after the equal sign... please help me to understand!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Just multiply top and bottom by e^{x}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2010
    Posts
    50
    but why to do that? could you explain
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    I'm not exactly sure myself. If I wanted to prove the identity, I would do this:

    \displaystyle{\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}=\frac{e^{x}-e^{-x}}{2}\cdot\frac{2}{e^{x}+e^{-x}}=\frac{\sinh(x)}{\cosh(x)}=\tanh(x).}

    It depends on how you define the hyperbolic tangent function, I guess. Make sense?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by problady View Post
    I dont have a clue how people come up to the second part of the equation after the equal sign... please help me to understand!
    \displaystyle\frac{e^x}{e^x}=1,

    one of the countless ways to express the value 1.

    Therefore...

    \displaystyle\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{e^x-e^{-x}}{e^x+e^{-x}}\, . \frac{e^x}{e^x}=\frac{e^xe^x-e^xe^{-x}}{e^xe^x+e^xe^{-x}}=\frac{e^{x+x}-e^{x-x}}{e^{x+x}+e^{x-x}}

    giving a reduced form of tanh(x)

    Hyperbolic function - Wikipedia, the free encyclopedia
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: April 6th 2010, 11:53 PM
  2. Hyperbolic Tangent
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 29th 2009, 07:33 PM
  3. hyperbolic tangent
    Posted in the Geometry Forum
    Replies: 3
    Last Post: September 26th 2009, 08:47 AM
  4. Tangent to hyperbolic function
    Posted in the Calculus Forum
    Replies: 4
    Last Post: June 13th 2009, 04:38 PM
  5. hyperbolic tangent issue
    Posted in the Calculus Forum
    Replies: 0
    Last Post: December 1st 2008, 06:30 AM

Search Tags


/mathhelpforum @mathhelpforum