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Thread: 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; Feb 10th 2011 at 11:56 AM. Reason: Re-titled.
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  2. #2
    MHF Contributor

    Apr 2005
    The "derivative", at a given point, of a function from $\displaystyle R^2$ to $\displaystyle R^2$, 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, $\displaystyle Dg= \begin{bmatrix}2x & 2y \\ -2y & 2x\end{bmatrix}$

    However, you say "f:reals^2 --> reals ^2" but then define "f(u,v) := (u^2) + (v^2)" which is $\displaystyle R^2$ to $\displaystyle R^1$. Assuming you meant to say "f:reals^2-->reals", then Df is a vector (again, more corrrectly, the linear transformation from $\displaystyle R^2$ to $\displaystyle R$ 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 $\displaystyle D(f\circ g)(x,y)$ is the product of those two matrices:
    $\displaystyle \begin{bmatrix}2(x^2- y^2) & 2(2xy)\end{bmatrix}\begin{bmatrix}2x & 2y \\ -2y & 2x\end{bmatrix}$
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