Jacobian for function with couple of couple

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 $$R^4$$,

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

• For $f$ why do you consider it as $f(x,y,u,v)$ anf not$f((x,y),(u,v))$ Commented Nov 20, 2018 at 21:07
• – smcc
Commented Nov 20, 2018 at 22:29