Results 1 to 6 of 6

Thread: 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.

    $\displaystyle 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
    12,880
    Thanks
    1946
    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.

    $\displaystyle 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.
    First of all this is very hard to read.

    I assume that it's

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


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

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

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


    Once you have those derivatives, multiply them together to get $\displaystyle \frac{dy}{dx}$ and then evaluate it at $\displaystyle x = 2$.
    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

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


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

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

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


    Once you have those derivatives, multiply them together to get $\displaystyle \frac{dy}{dx}$ and then evaluate it at $\displaystyle x = 2$.
    ok I expanded it and got $\displaystyle \frac{dy}{dx} = 2u-u^3$ 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
    12,880
    Thanks
    1946
    Quote Originally Posted by darksoulzero View Post
    ok I expanded it and got $\displaystyle \frac{dy}{dx} = 2u-u^3$ now how would I get the derivative in leibniz notation.
    No that doesn't look right.

    You have $\displaystyle y = 2u - u^3$.

    So now apply the power rule to each term to find $\displaystyle \frac{dy}{du}$.
    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 $\displaystyle y = 2u - u^3$.

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

    $\displaystyle =2u-u^3$
    $\displaystyle =2-3u^3$
    $\displaystyle =2-3(-x^{-2})^2$
    $\displaystyle =2-\frac{3}{-x^4}$
    $\displaystyle =2-\frac{3}{-(2)^4}$
    Last edited by darksoulzero; Mar 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
    12,880
    Thanks
    1946
    No.

    $\displaystyle y = 2u - u^3$


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

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

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


    $\displaystyle u = \frac{1}{x}$

    $\displaystyle = x^{-1}$.


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

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


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

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

    $\displaystyle = \frac{3}{x^4} - \frac{2}{x^2}$.
    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: Apr 23rd 2011, 07:51 AM
  2. Replies: 1
    Last Post: Mar 10th 2011, 02:23 AM
  3. Leibniz formula and chain rule
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Sep 26th 2010, 02:07 PM
  4. Leibniz notation.
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Apr 9th 2010, 04:20 AM
  5. Chain rule using Leibniz notation.
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Apr 2nd 2010, 01:34 PM

Search Tags


/mathhelpforum @mathhelpforum