Jacobian for function with couple of couple











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Caclulate the jacobian :



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



and



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



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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    up vote
    2
    down vote

    favorite












    Caclulate the jacobian :



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



    and



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



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










    share|cite|improve this question
























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Caclulate the jacobian :



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



      and



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



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










      share|cite|improve this question













      Caclulate the jacobian :



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



      and



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



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







      differential-equations multivariable-calculus partial-derivative






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      asked Nov 20 at 20:10









      Mary Maths

      165




      165






















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



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



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



          Then $$begin{align*}
          Df_1(x,y,u,v)&=begin{pmatrix}D_xf_1& D_yf_1 & D_uf_1 & D_vf_1end{pmatrix}=begin{pmatrix}u-3v & 0 & x & -3xend{pmatrix}\
          Df_2(x,y,u,v)&=begin{pmatrix}D_xf_2& D_yf_2 & D_uf_2 & D_vf_2end{pmatrix}=begin{pmatrix}0 & u & y & 0end{pmatrix}
          end{align*}$$



          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



          $$g(u,v)=detbegin{pmatrix}u_1 & v_1 \ u_2 & v_2end{pmatrix}=u_1v_2-u_2v_1$$



          so that $$Dg(u,v)=Dg(u_1,u_2,v_1,v_2)=begin{pmatrix}D_{u_1}g & D_{u_2}g & D_{v_1}g& D_{v_2}gend{pmatrix}=begin{pmatrix}v_2 & -v_1 & -u_2 & u_1end{pmatrix}$$






          share|cite|improve this answer





















          • For $ f $ why do you consider it as $f(x,y,u,v)$ anf not$f((x,y),(u,v))$
            – Mary Maths
            Nov 20 at 21:07










          • See here: math.stackexchange.com/questions/338319/…
            – smcc
            Nov 20 at 22:29











          Your Answer





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          1 Answer
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          up vote
          0
          down vote













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



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



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



          Then $$begin{align*}
          Df_1(x,y,u,v)&=begin{pmatrix}D_xf_1& D_yf_1 & D_uf_1 & D_vf_1end{pmatrix}=begin{pmatrix}u-3v & 0 & x & -3xend{pmatrix}\
          Df_2(x,y,u,v)&=begin{pmatrix}D_xf_2& D_yf_2 & D_uf_2 & D_vf_2end{pmatrix}=begin{pmatrix}0 & u & y & 0end{pmatrix}
          end{align*}$$



          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



          $$g(u,v)=detbegin{pmatrix}u_1 & v_1 \ u_2 & v_2end{pmatrix}=u_1v_2-u_2v_1$$



          so that $$Dg(u,v)=Dg(u_1,u_2,v_1,v_2)=begin{pmatrix}D_{u_1}g & D_{u_2}g & D_{v_1}g& D_{v_2}gend{pmatrix}=begin{pmatrix}v_2 & -v_1 & -u_2 & u_1end{pmatrix}$$






          share|cite|improve this answer





















          • For $ f $ why do you consider it as $f(x,y,u,v)$ anf not$f((x,y),(u,v))$
            – Mary Maths
            Nov 20 at 21:07










          • See here: math.stackexchange.com/questions/338319/…
            – smcc
            Nov 20 at 22:29















          up vote
          0
          down vote













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



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



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



          Then $$begin{align*}
          Df_1(x,y,u,v)&=begin{pmatrix}D_xf_1& D_yf_1 & D_uf_1 & D_vf_1end{pmatrix}=begin{pmatrix}u-3v & 0 & x & -3xend{pmatrix}\
          Df_2(x,y,u,v)&=begin{pmatrix}D_xf_2& D_yf_2 & D_uf_2 & D_vf_2end{pmatrix}=begin{pmatrix}0 & u & y & 0end{pmatrix}
          end{align*}$$



          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



          $$g(u,v)=detbegin{pmatrix}u_1 & v_1 \ u_2 & v_2end{pmatrix}=u_1v_2-u_2v_1$$



          so that $$Dg(u,v)=Dg(u_1,u_2,v_1,v_2)=begin{pmatrix}D_{u_1}g & D_{u_2}g & D_{v_1}g& D_{v_2}gend{pmatrix}=begin{pmatrix}v_2 & -v_1 & -u_2 & u_1end{pmatrix}$$






          share|cite|improve this answer





















          • For $ f $ why do you consider it as $f(x,y,u,v)$ anf not$f((x,y),(u,v))$
            – Mary Maths
            Nov 20 at 21:07










          • See here: math.stackexchange.com/questions/338319/…
            – smcc
            Nov 20 at 22:29













          up vote
          0
          down vote










          up vote
          0
          down vote









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



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



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



          Then $$begin{align*}
          Df_1(x,y,u,v)&=begin{pmatrix}D_xf_1& D_yf_1 & D_uf_1 & D_vf_1end{pmatrix}=begin{pmatrix}u-3v & 0 & x & -3xend{pmatrix}\
          Df_2(x,y,u,v)&=begin{pmatrix}D_xf_2& D_yf_2 & D_uf_2 & D_vf_2end{pmatrix}=begin{pmatrix}0 & u & y & 0end{pmatrix}
          end{align*}$$



          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



          $$g(u,v)=detbegin{pmatrix}u_1 & v_1 \ u_2 & v_2end{pmatrix}=u_1v_2-u_2v_1$$



          so that $$Dg(u,v)=Dg(u_1,u_2,v_1,v_2)=begin{pmatrix}D_{u_1}g & D_{u_2}g & D_{v_1}g& D_{v_2}gend{pmatrix}=begin{pmatrix}v_2 & -v_1 & -u_2 & u_1end{pmatrix}$$






          share|cite|improve this answer












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



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



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



          Then $$begin{align*}
          Df_1(x,y,u,v)&=begin{pmatrix}D_xf_1& D_yf_1 & D_uf_1 & D_vf_1end{pmatrix}=begin{pmatrix}u-3v & 0 & x & -3xend{pmatrix}\
          Df_2(x,y,u,v)&=begin{pmatrix}D_xf_2& D_yf_2 & D_uf_2 & D_vf_2end{pmatrix}=begin{pmatrix}0 & u & y & 0end{pmatrix}
          end{align*}$$



          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



          $$g(u,v)=detbegin{pmatrix}u_1 & v_1 \ u_2 & v_2end{pmatrix}=u_1v_2-u_2v_1$$



          so that $$Dg(u,v)=Dg(u_1,u_2,v_1,v_2)=begin{pmatrix}D_{u_1}g & D_{u_2}g & D_{v_1}g& D_{v_2}gend{pmatrix}=begin{pmatrix}v_2 & -v_1 & -u_2 & u_1end{pmatrix}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 20 at 20:54









          smcc

          4,297517




          4,297517












          • For $ f $ why do you consider it as $f(x,y,u,v)$ anf not$f((x,y),(u,v))$
            – Mary Maths
            Nov 20 at 21:07










          • See here: math.stackexchange.com/questions/338319/…
            – smcc
            Nov 20 at 22:29


















          • For $ f $ why do you consider it as $f(x,y,u,v)$ anf not$f((x,y),(u,v))$
            – Mary Maths
            Nov 20 at 21:07










          • See here: math.stackexchange.com/questions/338319/…
            – smcc
            Nov 20 at 22:29
















          For $ f $ why do you consider it as $f(x,y,u,v)$ anf not$f((x,y),(u,v))$
          – Mary Maths
          Nov 20 at 21:07




          For $ f $ why do you consider it as $f(x,y,u,v)$ anf not$f((x,y),(u,v))$
          – Mary Maths
          Nov 20 at 21:07












          See here: math.stackexchange.com/questions/338319/…
          – smcc
          Nov 20 at 22:29




          See here: math.stackexchange.com/questions/338319/…
          – smcc
          Nov 20 at 22:29


















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