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Math Help - Finding a Derivative

  1. #1
    Senior Member topher0805's Avatar
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    Finding a Derivative

    Differentiate.

    f(x) = \frac {7 - xe^x}{x + e^x}

    Sorry I am asking so many questions tonight, but I've been really sick for the past 10 days and have not been able to attend any classes. I'm trying to catch up, but some of these questions seem impossible.

    I thought I knew how to do this one. I used the quotient rule, f'(\frac {u}{v}) = \frac {vu' - uv'}{v^2} but I end up getting the wrong answer.

    These are my steps:

    \frac {(x+e^x)(7-xe^x)' - (7-xe^x)(x+e^x)'}{(x+e^x)^2}

    I then applied the sum and difference rules to get:

    \frac {(x+e^x)(7'-[xe^x]') - (7-xe^x)(x'+[e^x]')}{(x+e^x)^2}
    Last edited by topher0805; February 22nd 2008 at 07:32 PM.
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  2. #2
    Behold, the power of SARDINES!
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    f(x)=\frac{7-xe^x}{x+e^x}

    \frac{dy}{dx}=\frac{(x+e^x)(-xe^x-e^x)-(7-xe^x)(1+e^x)}{(x+e^x)^2}
    If you expand and collect like terms you get...

     \frac{dy}{dx}=\frac{-x^2e^x-e^{2x}-7e^{2x}-7}{(x+e^x)^2}
    Last edited by TheEmptySet; February 23rd 2008 at 06:46 AM.
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  3. #3
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    Quote Originally Posted by topher0805 View Post
    Differentiate.

    f(x) = \frac {7 - xe^x}{x + e^x}

    Sorry I am asking so many questions tonight, but I've been really sick for the past 10 days and have not been able to attend any classes. I'm trying to catch up, but some of these questions seem impossible.

    I thought I knew how to do this one. I used the quotient rule, f'(\frac {u}{v}) = \frac {vu' - uv'}{v^2} but I end up getting the wrong answer.

    These are my steps:

    \frac {(x+e^x)(7-xe^x)' - (7-xe^x)(x+e^x)'}{(x+e^x)^2}
    And a good step it is too. To continue:


    = \frac {(x+e^x)(-e^x - xe^x) - (7-xe^x)(1 + e^x)}{(x+e^x)^2}


    where to get (7 - x e^x)' I used the product rule to differentiate x e^x (note that the derivative of 7 is equal to zero). I hope getting (x+e^x)' is no bother for you .....


    = \frac {-(x+e^x)(e^x + xe^x) - (7-xe^x)(1 + e^x)}{(x+e^x)^2}


    Now it's simply a matter of expanding:


    = \frac {-x e^x -x^2 e^x - e^{2x} - x e^{2x} - 7 - 7 e^x + x e^x + x e^{2x}}{(x+e^x)^2}


    and simplifying:


    = \frac {-x^2 e^x - e^{2x} - 7 - 7 e^x}{(x+e^x)^2}.

    From here a variety of re-arrangements are possible .....

    (You should check in case I've made any careless mistakes).
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  4. #4
    Senior Member topher0805's Avatar
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    Smile

    Thanks for the help. I realized my mistake as I was typing out my steps.
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  5. #5
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    Quote Originally Posted by topher0805 View Post
    Thanks for the help. I realized my mistake as I was typing out my steps.
    You're welcome. Hope you're feeling much better. Good luck with it all.
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