Differential help please (2)

[nil101]

Please can you help simplify this one?

\(\displaystyle \L \b
y = \frac{{e^x }}{{e^x - e^{ - x} }}\)


\(\displaystyle \L u = e^x {\rm }u' = e^x\)

\(\displaystyle \L v = e^x - e^{ - x} {\rm }v' = e^x + e^{ - x}\)


\(\displaystyle \L \b
\frac{{dy}}{{dx}} = \frac{{e^x (e^x - e^{ - x} ) - e^x (e^x + e^{ - x} )}}{{(e^x + e^{ - x} )^2 }}\)

How do I simplify this ?

Thanks

 

[daon]

Please can you help simplify this one?

\(\displaystyle \L \b
y = \frac{{e^x }}{{e^x - e^{ - x} }}\)


\(\displaystyle \L u = e^x {\rm }u' = e^x\)

\(\displaystyle \L v = e^x - e^{ - x} {\rm }v' = e^x + e^{ - x}\)


\(\displaystyle \L \b
\frac{{dy}}{{dx}} = \frac{{e^x (e^x - e^{ - x} ) - e^x (e^x + e^{ - x} )}}{{(e^x + e^{ - x} )^2 }}\)

How do I simplify this ?

Thanks

Note that \(\displaystyle e^xe^{-x} = 1\). Also, I would just multiply out the top and see what cancels..

 

[galactus]

There are different methods.

Expanding the numerator we get:

\(\displaystyle \L\\(e^{2x}-1)-(e^{2x}+1)=-2\)

The denominator should be \(\displaystyle (e^{x}-e^{-x})^{2}=4(sinh(x))^{2}\)


Therefore, you have:

\(\displaystyle \L\\\frac{-1}{2(sinh(x))^{2}}\)

 

[nil101]

Thanks a lot guys.

 

[soroban]

Hello, nil101!

Here's one of the "different methods" that Galactus mentioned . . .

\(\displaystyle \L y\:=\:\frac{e^x}{e^x\,-\,e^{-x}}\)

Multiply top and bottom by \(\displaystyle e^x:\L\;\;y \:=\:\frac{e^{2x}}{e^{2x}\,-\,1}\)


We have: \(\displaystyle \L\:y \;= \;\frac{e^{2x}\,-\,1\,+\,1}{e^{2x}\,-\,1} \;= \;\frac{e^{2x}\,-\,1}{e^{2x}\,-\,1}\,+\,\frac{1}{e^{2x}\,-\,1}\)

\(\displaystyle \L\;\;\;=\;1\.+\.\frac{1}{e^{2x}\,-\,1}\;=\;1\,+\,(e^{2x}\,-\,1)^{-1}\)


Therefore: \(\displaystyle \L\:y'\;=\;0\,+\,(-1)(e^{2x}\,-\,1)^{-2}\cdot e^{2x}\cdot 2 \;= \;-\frac{2e^{2x}}{(e^{2x}\,-\,1)^2}\)