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Math Help - derivative

  1. #1
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    derivative

    find the derivative and simplify where possible:

    <br />
f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}}<br />

    when i do the quotient rule i get:

    <br />
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}<br />

    But i cant figure out what to do from here
    thanks : )
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  2. #2
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    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}}}.
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  3. #3
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    how did you combine the e^x with the e^{-x} to give you e^{2x}

    ....?
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  4. #4
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    Quote Originally Posted by Krizalid View Post
    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!
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  5. #5
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    Quote Originally Posted by qzno View Post
    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!
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  6. #6
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    still not making any sense.
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  7. #7
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    Quote Originally Posted by qzno View Post
    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}
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  8. #8
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    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
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  9. #9
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    Quote Originally Posted by qzno View Post
    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)}
    Last edited by Mush; January 14th 2009 at 02:32 PM.
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  10. #10
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    Thank you very much bud : )
    and im sorry about making you explain so much!
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  11. #11
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    Quote Originally Posted by qzno View Post
    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
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