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 $\cosh x + \sinh x$ "