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Math Help - [SOLVED] Tricky Diff question

  1. #1
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    Question [SOLVED] Tricky Diff question

    Hi - I just can't seem to get this one right. Looking to get the first derivative...

    (1 + xe^x)/(1 - xe^x)

    I assume that you use the quotient rule but I end up in a mess.

    Could someone please talk me thorugh this one?

    Thanks, D
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  2. #2
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    Hello dojo
    Quote Originally Posted by dojo View Post
    Hi - I just can't seem to get this one right. Looking to get the first derivative...

    (1 + xe^x)/(1 - xe^x)

    I assume that you use the quotient rule but I end up in a mess.

    Could someone please talk me thorugh this one?

    Thanks, D
    Yes, you'll need to use both the quotient and the product rules:
    f(x) = \frac{1+xe^x}{1-xe^x}

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

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

    = \frac{2e^x(x+1)}{(1-xe^x)^2}
    Does that look about right?

    Grandad

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  3. #3
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    The book gives the answer with the same numerator as you have bu the denom as (xe^2 - 1)^2

    Thanks for the workings - I think my algebra needs a bit of work. I think the book must be wrong as the quotien rule does state that v^2 is the denom.

    Thanks
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  4. #4
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    Hello dojo
    Quote Originally Posted by dojo View Post
    The book gives the answer with the same numerator as you have bu the denom as (xe^2 - 1)^2

    Thanks for the workings - I think my algebra needs a bit of work. I think the book must be wrong as the quotien rule does state that v^2 is the denom.

    Thanks
    Since (a-b)^2 is the same as (b-a)^2, you could certainly write the denominator as (xe^x-1)^2, but not (xe^2-1)^2.

    Are you sure that's what the book said? If so, it's just a typo.

    Grandad
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  5. #5
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    Thumbs up

    My typo!! thanks for the reply
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