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Let $f: \mathbb{R}^3 \to \mathbb{R}^2$ satisfy the conditions $f(0)=(1,2)$ and $$ Df(0)=\left[ \begin{array}{ccc} 1 & 2 & 3 \\ 0 & 0 & 1 \\ \end{array} \right] $$ Let $g: \mathbb{R}^2 \to \mathbb{R}^2$ be defined by the equation $$g(x,y)=(x+2y+1,3xy).$$ Find $D(gof)(0)$.

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  • $\begingroup$ Hint: What is $Dg(1,2)$? $\endgroup$ Oct 12, 2012 at 15:18

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Using the chain rule, you get $D(g\circ f)(0)={Dg}_{|_{f(0)}}\cdot Df(0)=Dg(1,2)\cdot Df(0)=\left[ \begin{array}{ccc} 1 & 2 \\ 3\cdot 2 & 3\cdot 1 \\ \end{array} \right]\cdot \left[ \begin{array}{ccc} 1 & 2 & 3 \\ 0 & 0 & 1 \\ \end{array} \right]=\left[ \begin{array}{ccc} 1 & 2 & 5 \\ 6 & 12 & 21 \\ \end{array} \right]$

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