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richyw
 3 years ago
Help! Vectorvalued functions.
define \(f:\mathbb{R}^2\rightarrow\mathbb{R}^3\) by \(f(u,v)=(u^25v,ve^{2u},2u\log(1+v^2))\). Suppose \(g:\mathbb{R}^2\rightarrow \mathbb{R}^2\) is of class \(C^1\), \(g(1,2)=(0,0)\) and \[Dg(1,2)=\left[\begin{matrix} 1 & 2 \\ 3 & 5\end{matrix}\right]\]Compute \(D(f\circ g)(1,2)\)
richyw
 3 years ago
Help! Vectorvalued functions. define \(f:\mathbb{R}^2\rightarrow\mathbb{R}^3\) by \(f(u,v)=(u^25v,ve^{2u},2u\log(1+v^2))\). Suppose \(g:\mathbb{R}^2\rightarrow \mathbb{R}^2\) is of class \(C^1\), \(g(1,2)=(0,0)\) and \[Dg(1,2)=\left[\begin{matrix} 1 & 2 \\ 3 & 5\end{matrix}\right]\]Compute \(D(f\circ g)(1,2)\)

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richyw
 3 years ago
Best ResponseYou've already chosen the best response.0so I know that\[Df=\left[\begin{matrix} 2u & 5 \\ 2ve^{2u} & e^{2u} \\ 2 & \frac{2v}{1+v^2}\end{matrix}\right]\]

richyw
 3 years ago
Best ResponseYou've already chosen the best response.0why wouldn't this just be \[Df(1,2)\cdot Dg(1,2)\]

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0If you were looking for \(D(f+g)(1,2)\) you could simplify as \(Df(1,2)+Dg(1,2)\), but differentiation is only linear with respect to addition. If you multiplied your functions together, you need to use the chain rule.
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