Results 1 to 10 of 10

Math Help - Second derivative

  1. #1
    Member
    Joined
    Dec 2008
    Posts
    92

    Second derivative

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

     f(x) = (x) / (x-1)

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

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

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

    Thanks in advance
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Dec 2008
    Posts
    51
    you can find the second derivative the same way u found the first, by using the quotient rule where  u=-1  <br />
and v=(x-1)^2

    hope that helps
    Last edited by iLikeMaths; March 24th 2009 at 09:21 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2009
    Posts
    53
    Quote Originally Posted by janedoe View Post
    I'm not sure how to find the second derivative of this:

     f(x) = (x) / (x-1)

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

    The answer for the 2nd derivative is:  (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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Dec 2008
    Posts
    92
    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  u=-1 <br />
and 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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Dec 2008
    Posts
    92
    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)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Dec 2008
    Posts
    51
    2x-2 is the same as 2(x-1) does that make any more sense.

    if u are not using the quotient rule, then what rule are you using
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Dec 2008
    Posts
    92
    Quote Originally Posted by iLikeMaths View Post
    2x-2 is the same as 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.....
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Dec 2008
    Posts
    51
    first \frac{-1}{(x-1)^2}derivative

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

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

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

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

    \frac{2}{(x-1)^3}
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Dec 2008
    Posts
    92
    Ahh thank you so much, I was forgetting the bottom half of the quotient rule...so stupid of me!!

    Thanks for being so helpful
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Dec 2008
    Posts
    51
    i thought so ..lol
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. contuous weak derivative $\Rightarrow$ classic derivative ?
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 22nd 2011, 03:37 AM
  2. Replies: 0
    Last Post: January 24th 2011, 12:40 PM
  3. [SOLVED] Definition of Derivative/Alt. form of the derivative
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 23rd 2010, 07:33 AM
  4. Derivative Increasing ==> Derivative Continuous
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: February 23rd 2010, 11:58 AM
  5. Replies: 2
    Last Post: November 6th 2009, 03:51 PM

Search Tags


/mathhelpforum @mathhelpforum