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Math Help - how to dolce this?deviative of exponenetial functions

  1. #1
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    how to do this?deviative of exponenetial functions

    <br />
\frac{x^2 -1}{ e^2^x(x+1)}<br /> <br />

    the outcome supposs to be like this

    <br />
\frac{2x-3}{e^2^x}<br /> <br />

    i dont even know which part shld i deviate first

    i am trying to learn the steps to solving this pls help me
    Last edited by neodeath; February 19th 2006 at 08:31 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by neodeath
    <br />
\frac{x^2 -1}{ e^2^x(x+1)}<br /> <br />

    the outcome supposs to be like this

    <br />
\frac{2x-3}{e^2^x}<br /> <br />

    i dont even know which part shld i deviate first

    i am trying to learn the steps to solving this pls help me
    <br />
\frac{x^2 -1}{ e^2^x(x+1)}=\frac{(x+1)(x-1)}{e^{2x}(x+1)}=\frac{x-1}{e^{2x}}<br />

    Which can be differentiated using the quotient rule, or if you are like
    me and can't be bothered with remembering that, we can rewrite this as:

    <br />
\frac{d}{dx}\ \frac{x-1}{e^{2x}}=\frac{d}{dx}(x-1)e^{-2x}<br />

    which can be done using the product rule:

    <br />
\frac{d}{dx}(x-1)e^{-2x}=e^{-2x}+(x-1)(-2)e^{-2x}=\frac{3-2x}{e^{2x}}<br />

    RonL
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  3. #3
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    wow ddnt know cna use product and quotion rule

    by the way i made some mistakes too the ans supposs to be like this:

    <br />
-\frac{2x-3}{e^2^x}<br />

    seriosuly embaress >.<

    but any way so to do question like this i shld simplify them then deviate them rite?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by neodeath
    wow ddnt know cna use product and quotion rule

    by the way i made some mistakes too the ans supposs to be like this:

    <br />
-\frac{2x-3}{e^2^x}<br />

    seriosuly embaress >.<

    but any way so to do question like this i shld simplify them then deviate them rite?
    Yes (note the answer that I got is the same as the one you are now quoting)

    RonL
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