Write the given hyperbolic function as simply as possible

[chwala]

Homework Statement

##\dfrac{e^x}{1+e^{2x}}##

Relevant Equations

hyperbolic equations

My take;

##2\cosh x = e^x +e^{-x}##

I noted that i could multiply both sides by ##e^x## i.e

##e^x⋅2\cosh x = e^x(e^x +e^{-x})##

##e^x⋅2\cosh x = e^{2x}+1##

thus,

##\dfrac{e^x}{1+e^{2x}}=\dfrac{\cosh x + \sinh x}{e^x⋅2\cosh x}##

##= \dfrac{\cosh x + \sinh x}{(\cosh x + \sinh x)⋅2\cosh x}##

##=\dfrac{1}{2\cosh x}##
any other approach is welcome.

 

[BvU]

Homework Statement:: ##\dfrac{e^x}{1+e^{2x}}##
Relevant Equations:: hyperbolic equations

both sides

There are no 'sides'
There is no 'equation'

You post an expression. If you divide numerator and denominator by ##e^x## you see that you can rewrite the expression as ##1\over 2\cosh x##: the numerator is now ##1## and the denominator is now ##2\cosh x##. There is no need to write ##1## as ##\cosh x + \sinh x##

Cheers !

##\ ##

 

[chwala]

There are no 'sides'
There is no 'equation'

You post an expression. If you divide numerator and denominator by ##e^x## you see that you can rewrite the expression as ##1\over 2\cosh x##: the numerator is now ##1## and the denominator is now ##2\cosh x##. There is no need to write ##1## as ##\cosh x + \sinh x##

Cheers !

##\ ##

...seen that...correct man ! it's an expression ...i just posted exactly as it appears on textbook...i should have checked that or rather introduced ##f(x)## on the lhs.

 

[BvU]

There is no need to write ##1## as ##\cosh x + \sinh x##

the more so because it is totally incorrect ! My bad, I should have written "##e^x## as ##\cosh x + \sinh x## "