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Thread: Second derivative

  1. #1
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    Second derivative

    I'm not sure how to find the second derivative of this:

    $\displaystyle f(x) = (x) / (x-1)$

    For the 1st derivative I got: $\displaystyle (-1) / [ (x-1)^2 ]$

    The answer for the 2nd derivative is: $\displaystyle (2) / [ (x-1)^3 ] $

    But I can't see how they got it...

    Thanks in advance
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  2. #2
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    you can find the second derivative the same way u found the first, by using the quotient rule where $\displaystyle u=-1
    $ and $\displaystyle v=(x-1)^2$

    hope that helps
    Last edited by iLikeMaths; Mar 24th 2009 at 08:21 AM.
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  3. #3
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    Quote Originally Posted by janedoe View Post
    I'm not sure how to find the second derivative of this:

    $\displaystyle f(x) = (x) / (x-1)$

    For the 1st derivative I got: $\displaystyle (-1) / [ (x-1)^2 ]$

    The answer for the 2nd derivative is: $\displaystyle (2) / [ (x-1)^3 ] $

    But I can't see how they got it...

    Thanks in advance
    (-1)/(x-1)^2

    So, you do it in the same way as the first. Quotient rule, and then you've got a chain rule in the denominator.

    alternatively, you can bring the whole bottom up to the top,

    -1[(x-1)^-2]

    and just do the chain rule, so

    -1(-2)[(x-1)^-3] =

    2/(x-1)^3
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  4. #4
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    Quote Originally Posted by iLikeMaths View Post
    you can find the second derivative the same way u found the first, by using the quotient rule where $\displaystyle u=-1
    $ and $\displaystyle v=(x-1)^2$

    hope that helps
    Well I tried that, and I got 2x-2 for the second derivative. However, the answer is not that. Where am I going wrong..I used the quotient rule like you said which makes sense.
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  5. #5
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    Quote Originally Posted by coolguy99 View Post
    (-1)/(x-1)^2

    So, you do it in the same way as the first. Quotient rule, and then you've got a chain rule in the denominator.

    alternatively, you can bring the whole bottom up to the top,

    -1[(x-1)^-2]

    and just do the chain rule, so

    -1(-2)[(x-1)^-3] =

    2/(x-1)^3

    We haven't learned this yet, so I would like to know how to do it without this rule (for ex. using the quotient rule, because I used the q.r. and got the wrong answer)
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  6. #6
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    $\displaystyle 2x-2$ is the same as $\displaystyle 2(x-1)$ does that make any more sense.

    if u are not using the quotient rule, then what rule are you using
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  7. #7
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    Quote Originally Posted by iLikeMaths View Post
    $\displaystyle 2x-2$ is the same as $\displaystyle 2(x-1)$ does that make any more sense.

    if u are not using the quotient rule, then what rule are you using
    Yes I am using the quotient rule, but when I did I got 2x-2...yes that is the same as 2(x-1) but the answer is 2/(x-1)^3.....
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  8. #8
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    first $\displaystyle \frac{-1}{(x-1)^2}$derivative

    using the quotient rule $\displaystyle u=-1$ and $\displaystyle v=(x-2)^2$ and $\displaystyle \frac{du}{dx}=0$ and $\displaystyle \frac{dv}{dx}=2(x-1)$ and $\displaystyle v^2 =(x-1)^4$

    $\displaystyle \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$

    $\displaystyle \frac{0 -(-2(x-1))}{(x-1)^4}$

    $\displaystyle \frac{2(x-1)}{(x-1)^4}$

    $\displaystyle \frac{2}{(x-1)^3}$
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  9. #9
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    Ahh thank you so much, I was forgetting the bottom half of the quotient rule...so stupid of me!!

    Thanks for being so helpful
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  10. #10
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    i thought so ..lol
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