Results 1 to 3 of 3

Math Help - Tricky derivative

  1. #1
    Senior Member DivideBy0's Avatar
    Joined
    Mar 2007
    From
    Melbourne, Australia
    Posts
    432

    Tricky derivative

    I need help to do:

    y=\sqrt[x]{\frac{1-x}{1+x}}

    The answer says
    \frac{1-x-x^2}{(1+x)(1-x^2)^{\frac{1}{2}}}, but I don't think that is correct

    Thanks
    Last edited by DivideBy0; August 12th 2007 at 12:36 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,715
    Thanks
    634
    Hello, DivideBy0!

    I assume we are to differentiate the fuction . . .


    y\:=\:\sqrt[x]{\frac{1-x}{1+x}}
    The answer says: . \frac{1-x-x^2}{(1+x) (1-x^2)^{\frac{1}{2}}} .but I don't think that is correct
    Please check the problem for typos . . . Something is terribly wrong!

    If that is really an x^{th} root, the answer cannot have a square root in it.

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,902
    Thanks
    329
    Awards
    1
    Quote Originally Posted by DivideBy0 View Post
    I need help to do:

    y=\sqrt[x]{\frac{1-x}{1+x}}

    The answer says
    \frac{1-x-x^2}{(1+x)(1-x^2)^{\frac{1}{2}}}, but I don't think that is correct

    Thanks
    You are correct. It isn't right.

    The derivative of a^x is ln(a) \cdot a^x, so using the chain rule on
    y=\sqrt[x]{\frac{1-x}{1+x}} = \left ( \frac{1 - x}{1 + x} \right ) ^{1/x}
    we obtain:
    y^{\prime} = ln \left ( \frac{1 - x}{1 + x} \right ) \cdot \left ( \frac{1 - x}{1 + x} \right ) ^{1/x} \cdot \left ( - \frac{1}{x^2} \right ) \cdot \left [ \frac{1 - x}{1 + x} \right ]^{\prime}

    y^{\prime} = ln \left ( \frac{1 - x}{1 + x} \right ) \cdot \left ( \frac{1 - x}{1 + x} \right ) ^{1/x} \cdot \left ( - \frac{1}{x^2} \right ) \cdot \left ( \frac{-2}{(1 + x)^2} \right )

    which I would simply leave as:
    y^{\prime} = \left ( \frac{2}{x^2(x + 1)^2} \right ) \cdot \left ( \frac{1 - x}{1 + x} \right ) ^{1/x} \cdot ln \left ( \frac{1 - x}{1 + x} \right )

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Tricky Derivative Question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 22nd 2011, 09:40 PM
  2. tricky derivative?
    Posted in the Calculus Forum
    Replies: 7
    Last Post: February 25th 2010, 11:46 AM
  3. Tricky Derivative
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 15th 2009, 04:23 AM
  4. Tricky Derivative
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 12th 2009, 10:31 AM
  5. Derivative, really tricky one!! Am I right???
    Posted in the Calculus Forum
    Replies: 6
    Last Post: October 12th 2008, 08:52 PM

Search Tags


/mathhelpforum @mathhelpforum