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Let $f: \mathbb{R}^3 \to \mathbb{R}^2$ satisfy the condition $f(0)=(1,2)$ and $$Df(0)= \left( {\begin{array}{cc} 1 & 2 & 3 \\ 0 & 1 & 1 \end{array} } \right) $$

Let $g : \mathbb{R}^2 \to \mathbb{R}^2$ be defined by $g(x,y)=(x+2y+1,3xy)$.

My question is how can I find $D(g \circ f)(0)$?

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    $\begingroup$ Using this notation, the chain rule is really easy: $D(g\circ f) = D(g)\circ D(f)$. $\endgroup$
    – Arthur
    Mar 14, 2015 at 10:41
  • $\begingroup$ @Arthur i tried this but I was unsure about the calculations, can you please show me? $\endgroup$
    – Lori
    Mar 14, 2015 at 13:02

1 Answer 1

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\begin{align} f(x,y,z) &= (u(x,y,z),v(x,y,z))\\ f(0,0,0) &= (1,2)\\ g(u,v) &= (u+2v+1,3uv) \end{align}\

\begin{align} Dg(u,v) &= \left( {\begin{array}{*{20}{c}} {{\partial _u}(u + 2v + 1)}&{{\partial _v}(u + 2v + 1)}\\ {{\partial _u}(3uv)}&{{\partial _v}(3uv)} \end{array}} \right) =\left( {\begin{array}{*{20}{c}} 1&2\\ {3v}&{3u} \end{array}} \right) \end{align}\

$$ Dg(f(0)) = Dg(1,2) = \left( {\begin{array}{*{20}{c}} 1&2\\ {3 \cdot 2}&{3 \cdot 1} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&2\\ 6&3 \end{array}} \right)$$

$$Df(0) = \left( {\begin{array}{*{20}{c}} 1&2&3\\ 0&1&1 \end{array}} \right)$$\

$$D(g \circ f) = Dg \circ Df$$

$$Dg(f(0))Df(0) = \left( {\begin{array}{*{20}{c}} 1&2\\ 6&3 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 1&2&3\\ 0&1&1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&4&5\\ 6&{15}&{21} \end{array}} \right)$$

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  • $\begingroup$ Why is $f$ defined in each component with the same $u$? $\endgroup$
    – Lori
    Mar 14, 2015 at 21:26
  • $\begingroup$ @Lori: Second u must be v. It's a typo. Thank you. Just edited. $\endgroup$
    – Frieder
    Mar 14, 2015 at 21:37

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