Math Help - derivative

1. 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 : )

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

3. how did you combine the $e^x$ with the $e^{-x}$ to give you $e^{2x}$

....?

4. 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!

5. 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!

6. still not making any sense.

7. Originally Posted by qzno
still not making any sense.

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

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

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

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

8. 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

9. 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: $f'(x)= \frac{h(x)g'(x) - g(x)h'(x)}{(h(x))^2}$

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

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

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

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

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

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

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

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

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

10. Thank you very much bud : )
and im sorry about making you explain so much!

11. 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