# derivative

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• Jan 14th 2009, 02:31 PM
qzno
derivative
find the derivative and simplify where possible:

$
f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}}
$

when i do the quotient rule i get:

$
f'(x) = \frac{(e^x + e^{-x})(e^x - e^{-x})(-1) - (e^x - e^{-x})(e^x + e^{-x})(-1)}{(e^x + e^{-x})^2}
$

But i cant figure out what to do from here
thanks : )
• Jan 14th 2009, 02:45 PM
Krizalid
$f(x)=\frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}=\frac{{{e}^{2x}}-1}{{{e}^{2x}}+1}=\frac{{{e}^{2x}}+1-2}{{{e}^{2x}}+1}=1-\frac{2}{{{e}^{2x}}+1},$ hence $f'(x)=\frac{4{{e}^{2x}}}{{{\left( {{e}^{2x}}+1 \right)}^{2}}}=\frac{4}{{{\left( {{e}^{x}}+{{e}^{-x}} \right)}^{2}}}.$
• Jan 14th 2009, 02:49 PM
qzno
how did you combine the $e^x$ with the $e^{-x}$ to give you $e^{2x}$

....?
• Jan 14th 2009, 02:51 PM
Mush
Quote:

Originally Posted by Krizalid
$f(x)=\frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}=\frac{{{e}^{2x}}-1}{{{e}^{2x}}+1}=\frac{{{e}^{2x}}+1-2}{{{e}^{2x}}+1}=1-\frac{2}{{{e}^{2x}}+1},$ hence $f'(x)=\frac{4{{e}^{2x}}}{{{\left( {{e}^{2x}}+1 \right)}^{2}}}=\frac{4}{{{\left( {{e}^{x}}+{{e}^{-x}} \right)}^{2}}}.$

$= ((\frac{e^x+e^{-x}}{2})^2)^{-1}$

$= (\cosh^2{(x)})^{-1}$

$= \frac{1}{\cosh^2{(x)}}$

$= \text{sech}^2{(x)}$

Which is expected given that your $f(x) = \tanh{(x)}$ if you look carefully enough!
• Jan 14th 2009, 02:52 PM
Mush
Quote:

Originally Posted by qzno
how did you combine the $e^x$ with the $e^{-x}$ to give you $e^{2x}$

....?

Take out a factor of $e^{-x}$ on top and bottom of the fraction!
• Jan 14th 2009, 03:01 PM
qzno
still not making any sense.
• Jan 14th 2009, 03:04 PM
Mush
Quote:

Originally Posted by qzno
still not making any sense.

$\displaystyle \frac{e^x-e^{-x}}{e^x+e^{-x}} = \frac{e^{-x}(\frac{e^{x}}{e^{-x}} - \frac{e^{-x}}{e^{-x}})}{e^{-x}(\frac{e^{x}}{e^{-x}} + \frac{e^{-x}}{e^{-x}})}$

$\displaystyle = \frac{e^{x}.e^{x} - 1}{e^{x}.e^{x} + 1}$

$\displaystyle = \frac{e^{x+x} - 1}{e^{x+x} + 1}$

$\displaystyle = \frac{e^{2x} - 1}{e^{2x} + 1}$
• Jan 14th 2009, 03:07 PM
qzno
can someone do it with the quotient rule please
because thats how ive been learning is via the three rules

1. power rule
2. product rule
3. quotient rule
• Jan 14th 2009, 03:12 PM
Mush
Quote:

Originally Posted by qzno
can someone do it with the quotient rule please
because thats how ive been learning is via the three rules

1. power rule
2. product rule
3. quotient rule

Okay let $g(x) = e^x -e^{-x}$ and $h(x)= e^x +e^{-x}$

Then:

$g'(x) = e^x -e^{-x}(-1) = e^x+e^{-x}$

and

$h'(x) = e^x + e^{-x}(-1) = e^x-e^{-x}$

Now, $f(x) = \frac{g(x)}{h(x)}$

Hence, by the quotient rule: $f'(x)= \frac{h(x)g'(x) - g(x)h'(x)}{(h(x))^2}$

$= \frac{(e^x +e^{-x})(e^x+e^{-x}) - (e^x -e^{-x})(e^x-e^{-x})}{(e^x +e^{-x})^2}$

$= \frac{(e^x.e^x+e^x.e^{-x} +e^{-x}.e^x+e^{-x}e^{-x}) - (e^x.e^x-e^x.e^{-x} -e^{-x}.e^x+e^{-x}e^{-x})}{(e^x +e^{-x})^2}$

$= \frac{(e^{2x}+e^0 +e^0+e^{-2x}) - (e^{2x}-e^0 -e^0+e^{-2x})}{(e^x +e^{-x})^2}$

$= \frac{e^{2x}+1 +1+e^{-2x} - e^{2x}+1+1-e^{-2x}}{(e^x +e^{-x})^2}$

$= \frac{4}{(e^x +e^{-x})^2}$

$= ((\frac{e^x+e^{-x}}{2})^2)^{-1}$

$= (\cosh^2{(x)})^{-1}$

$= \frac{1}{\cosh^2{(x)}}$

$= \text{sech}^2{(x)}$
• Jan 14th 2009, 03:16 PM
qzno
Thank you very much bud : )
and im sorry about making you explain so much!
• Jan 14th 2009, 03:24 PM
Mush
Quote:

Originally Posted by qzno
Thank you very much bud : )
and im sorry about making you explain so much!

No problem. View the edits to see it in its final form! Took me a wee while :P