Discussion
Let \(f : \mathbf{R}^{3} \to \mathbf{R}^{2}\) be defined by the equation \(f(x, y, z) = (xy, y + z)\) and \(g : \mathbf{R}^{2} \to \mathbf{R}^{2}\) be defined as \(g(x, y) = (x + y, y)\). Then the derivative \(D(g \circ f)(0, 0, 0) =\)
Answer: C
\(D(g \circ f)(0, 0, 0)\) is \(\begin{bmatrix} 0 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix}\).
Compose first: \(g(f(x, y, z)) = (xy + y + z,\ y + z)\). Its Jacobian is
\[ \begin{bmatrix} y & x + 1 & 1 \\ 0 & 1 & 1 \end{bmatrix} \]
which at the origin is \(\begin{bmatrix} 0 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix}\). The chain rule agrees: \(Dg = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\) and \(Df(0) = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 1 \end{bmatrix}\), whose product is the same matrix.
The \(3 \times 2\) option has the wrong shape for a map from \(\mathbf{R}^{3}\) to \(\mathbf{R}^{2}\), and the other two put non-zero entries in the first column, but nothing depends on \(x\) to first order at the origin.
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