Results 1 to 6 of 6

Math Help - Chain Rule Leibniz Notation

  1. #1
    Junior Member
    Joined
    Nov 2009
    Posts
    46

    Chain Rule Leibniz Notation

    Hi, I have been stuck on this question for a really long time any help would be appreciated.

    Using leibniz notation, apply the chain to determine (dy/dx) at the indicated value of x.

    y= u(2-u^2), u= (1/x), x=2


    I think that I have to use the product rule and then thee chain rule but I'm lost.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,560
    Thanks
    1425
    Quote Originally Posted by darksoulzero View Post
    Hi, I have been stuck on this question for a really long time any help would be appreciated.

    Using leibniz notation, apply the chain to determine (dy/dx) at the indicated value of x.

    y= <i>u</i>(2-<i>(u^2)</i>), <i>u</i>= (1/x), x=2


    I think that I have to use the product rule and then thee chain rule but I'm lost.
    First of all this is very hard to read.

    I assume that it's

    y = u(2 - u^2) and u = \frac{1}{x}, where x = 2.


    You need to find \frac{du}{dx} and \frac{dy}{du}.

    You are correct that to find \frac{dy}{du} you should use the product rule. Another alternative is to expand the expression first.

    You should also note that \frac{1}{x} = x^{-1}.


    Once you have those derivatives, multiply them together to get \frac{dy}{dx} and then evaluate it at x = 2.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2009
    Posts
    46
    Quote Originally Posted by Prove It View Post
    First of all this is very hard to read.

    I assume that it's

    y = u(2 - u^2) and u = \frac{1}{x}, where x = 2.


    You need to find \frac{du}{dx} and \frac{dy}{du}.

    You are correct that to find \frac{dy}{du} you should use the product rule. Another alternative is to expand the expression first.

    You should also note that \frac{1}{x} = x^{-1}.


    Once you have those derivatives, multiply them together to get \frac{dy}{dx} and then evaluate it at x = 2.
    ok I expanded it and got \frac{dy}{dx} = 2u-u^3 now how would I get the derivative in leibniz notation.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,560
    Thanks
    1425
    Quote Originally Posted by darksoulzero View Post
    ok I expanded it and got \frac{dy}{dx} = 2u-u^3 now how would I get the derivative in leibniz notation.
    No that doesn't look right.

    You have y = 2u - u^3.

    So now apply the power rule to each term to find \frac{dy}{du}.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Nov 2009
    Posts
    46
    Quote Originally Posted by Prove It View Post
    No that doesn't look right.

    You have y = 2u - u^3.

    So now apply the power rule to each term to find \frac{dy}{du}.
    ok I think I got it.

    =2u-u^3
    =2-3u^3
    =2-3(-x^{-2})^2
    =2-\frac{3}{-x^4}
    =2-\frac{3}{-(2)^4}
    Last edited by darksoulzero; March 7th 2010 at 06:00 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,560
    Thanks
    1425
    No.

    y = 2u - u^3


    \frac{dy}{du} = 2 - 3u^2

     = 2 - 3\left(\frac{1}{x}\right)^2

     = 2 - \frac{3}{x^2}.


    u = \frac{1}{x}

     = x^{-1}.


    \frac{du}{dx} = -x^{-2}

     = -\frac{1}{x^2}.


    \frac{dy}{dx} = \frac{du}{dx}\,\frac{dy}{du}

     = -\frac{1}{x^2}\left(2 - \frac{3}{x^2}\right)

     = \frac{3}{x^4} - \frac{2}{x^2}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. leibniz notation
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 23rd 2011, 07:51 AM
  2. Replies: 1
    Last Post: March 10th 2011, 02:23 AM
  3. Leibniz formula and chain rule
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 26th 2010, 02:07 PM
  4. Leibniz notation.
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 9th 2010, 04:20 AM
  5. Chain rule using Leibniz notation.
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 2nd 2010, 01:34 PM

Search Tags


/mathhelpforum @mathhelpforum