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

Printable View

• September 2nd 2010, 08:16 AM
problady
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!
• September 2nd 2010, 08:21 AM
Ackbeet
Just multiply top and bottom by $e^{x}$.
• September 2nd 2010, 11:05 AM
problady
but why to do that? could you explain
• September 2nd 2010, 11:25 AM
Ackbeet
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?
• September 2nd 2010, 11:54 AM
Archie Meade
Quote:

Originally Posted by problady
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