Results 1 to 2 of 2

Math Help - D(f o g))(x,y) by the chain rule.

  1. #1
    Feb 2011

    D(f o g))(x,y) by the chain rule.

    if we have g: reals^2 -- reals ^2 and f: reals^2 -- reals ^2
    And g(x,y) :=((x^2) (y^2), 2xy) f(u,v) := (u^2) + (v^2)

    How do we evaluate (D(f o g))(x,y) by the chain rule.
    Last edited by mr fantastic; February 10th 2011 at 11:56 AM. Reason: Re-titled.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Apr 2005
    The "derivative", at a given point, of a function from R^2 to R^2, at a given point, is a two by two matrix (more correctly, the linear function given by that matrix in the standard basis).

    In particular, Dg= \begin{bmatrix}2x & 2y \\ -2y & 2x\end{bmatrix}

    However, you say "f:reals^2 --> reals ^2" but then define "f(u,v) := (u^2) + (v^2)" which is R^2 to R^1. Assuming you meant to say "f:reals^2-->reals", then Df is a vector (again, more corrrectly, the linear transformation from R^2 to R defined by taking the dot product of that vector with a vector v.). Here, that would be [tex]\begin{bmatrix}2u & 2v\end{bmatrix}.

    The chain rule says that D(f\circ g)(x,y) is the product of those two matrices:
    \begin{bmatrix}2(x^2- y^2) & 2(2xy)\end{bmatrix}\begin{bmatrix}2x & 2y \\ -2y & 2x\end{bmatrix}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 9
    Last Post: November 9th 2010, 01:40 AM
  2. Chain Rule Inside of Chain Rule
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 22nd 2009, 08:50 PM
  3. Replies: 5
    Last Post: October 19th 2009, 01:04 PM
  4. Replies: 3
    Last Post: May 25th 2009, 06:15 AM
  5. Replies: 2
    Last Post: December 13th 2007, 05:14 AM

Search Tags

/mathhelpforum @mathhelpforum