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Caclulate the jacobian :

$ f:\mathbb{R}^2\times\mathbb{R}^2\to\mathbb{R}^2 \\ ((x,y),(u,v))\to(ux-3xv,yu) $

and

$g:\mathbb{R}^2\times\mathbb{R}^2\to\mathbb{R}\\ (U,V)\to det(u,v)$

for $f$ I don't how to differantiate for a couple of couple. Can I consider it as map form $R^4$,

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1 Answer 1

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In the first one you have $f=(f_1,f_2)$ where $f_1:\mathbb{R}^4\to\mathbb{R}$ is given by $$f_1(x,y,u,v)=ux-3xv$$

and $f_2:\mathbb{R}^4\to \mathbb{R}$ is given by

$$f_2(x,y,u,v)=yu.$$

Then $$\begin{align*} Df_1(x,y,u,v)&=\begin{pmatrix}D_xf_1& D_yf_1 & D_uf_1 & D_vf_1\end{pmatrix}=\begin{pmatrix}u-3v & 0 & x & -3x\end{pmatrix}\\ Df_2(x,y,u,v)&=\begin{pmatrix}D_xf_2& D_yf_2 & D_uf_2 & D_vf_2\end{pmatrix}=\begin{pmatrix}0 & u & y & 0\end{pmatrix} \end{align*}$$

So $$Df(x,y,u,v)=\begin{pmatrix}u-3v & 0 & x & -3x\\0 & u & y & 0 \end{pmatrix}$$

For the second one, denote $u=(u_1,u_2)$ and $v=(v_1,v_2)$. Then

$$g(u,v)=det\begin{pmatrix}u_1 & v_1 \\ u_2 & v_2\end{pmatrix}=u_1v_2-u_2v_1$$

so that $$Dg(u,v)=Dg(u_1,u_2,v_1,v_2)=\begin{pmatrix}D_{u_1}g & D_{u_2}g & D_{v_1}g& D_{v_2}g\end{pmatrix}=\begin{pmatrix}v_2 & -v_1 & -u_2 & u_1\end{pmatrix}$$

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