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Math Help - Use of chain rule for two variables

  1. #1
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    Use of chain rule for two variables

    Let \underline{g}:\mathbb{R}^2 \rightarrow \mathbb{R}^2 be defined by \underline{g}(x,y)=(x^2-y^2, 2xy).
    Let f:\mathbb{R}^2 \rightarrow \mathbb{R} be defined by f(u,v)= u^2+v^2
    Evaluate (D(f \circ \underline{g}))(x,y)

    My attempt:

    Chain rule for two variables: (D(f \circ \underline{g}))(x,y)=Df(\underline{g}(x,y)).D\unde  rline{g}(x,y)

    Df(\underline{g}(x,y))=\begin{pmatrix} \frac{\partial (x^2+y^2)^2}{\partial (x^2-y^2)} & \frac{\partial (x^2+y^2)^2}{\partial (2xy)}\end{pmatrix}

    D\underline{g}(\underline{x})= \begin{pmatrix} \frac{\partial (x^2-y^2)}{\partial (x)} & \frac{\partial (x^2-y^2)}{\partial (y)} \\ \frac{\partial (2xy)}{\partial (x)} & \frac{\partial (2xy)}{\partial (y}\end{pmatrix}=\begin{pmatrix} 2x & -2y \\ 2y & 2x \end{pmatrix}

    My problem is that i have no idea how to differentiate the first matrix???
    Also the matrix dimensions dont work for matrix multiplication
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  2. #2
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    Quote Originally Posted by FGT12 View Post
    Let \underline{g}:\mathbb{R}^2 \rightarrow \mathbb{R}^2 be defined by \underline{g}(x,y)=(x^2-y^2, 2xy).
    Let f:\mathbb{R}^2 \rightarrow \mathbb{R} be defined by f(u,v)= u^2+v^2
    Evaluate (D(f \circ \underline{g}))(x,y)

    My attempt:

    Chain rule for two variables: (D(f \circ \underline{g}))(x,y)=Df(\underline{g}(x,y)).D\unde  rline{g}(x,y)

    Df(\underline{g}(x,y))=\begin{pmatrix} \frac{\partial (x^2+y^2)^2}{\partial (x^2-y^2)} & \frac{\partial (x^2+y^2)^2}{\partial (2xy)}\end{pmatrix}

    D\underline{g}(\underline{x})= \begin{pmatrix} \frac{\partial (x^2-y^2)}{\partial (x)} & \frac{\partial (x^2-y^2)}{\partial (y)} \\ \frac{\partial (2xy)}{\partial (x)} & \frac{\partial (2xy)}{\partial (y}\end{pmatrix}=\begin{pmatrix} 2x & -2y \\ 2y & 2x \end{pmatrix}

    My problem is that i have no idea how to differentiate the first matrix???
    Also the matrix dimensions dont work for matrix multiplication
    If you write (u,v) = \underline{g}(x,y), then Df(u,v) = \begin{pmatrix}\frac{\partial (u^2+v^2)}{\partial u} & \frac{\partial (u^2+v^2)}{\partial v}\end{pmatrix} = \bigl(2u\ \ 2v\bigr) = \bigl(2(x^2-y^2)\ \ 4xy\bigr). So Df(\underline{g}(x,y))= \bigl(2(x^2-y^2)\ \ 4xy\bigr) \begin{pmatrix} 2x & -2y \\ 2y & 2x \end{pmatrix}, which is a 1x2 matrix, as it should be for the derivative of the function  f \circ \underline{g}: \mathbb{R}^2 \rightarrow \mathbb{R}.
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