Derivative of a function defined in terms of another function












0












$begingroup$


Let $F: mathbf{R^2} rightarrow mathbf{R}$ have continuous second order partial derivatives. Assume its gradient is $nabla f(0,0) = (1,2)$ and its Hessian matrix is
$$begin{pmatrix}
1 & 2 \
2 &4
end{pmatrix} = nabla ^2 f $$



Define $phi (t) = f(t, 2t)$. Find $phi ^ prime (0)$ and $phi ^ {prime prime} (0)$.



Any hints? I know how to do this when $f$ is explicitly defined, but not when just given the gradient/Hessian at a certain point.






real-analysis







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    0












    $begingroup$


    Let $F: mathbf{R^2} rightarrow mathbf{R}$ have continuous second order partial derivatives. Assume its gradient is $nabla f(0,0) = (1,2)$ and its Hessian matrix is
    $$begin{pmatrix}
    1 & 2 \
    2 &4
    end{pmatrix} = nabla ^2 f $$



    Define $phi (t) = f(t, 2t)$. Find $phi ^ prime (0)$ and $phi ^ {prime prime} (0)$.



    Any hints? I know how to do this when $f$ is explicitly defined, but not when just given the gradient/Hessian at a certain point.






    real-analysis







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $F: mathbf{R^2} rightarrow mathbf{R}$ have continuous second order partial derivatives. Assume its gradient is $nabla f(0,0) = (1,2)$ and its Hessian matrix is
      $$begin{pmatrix}
      1 & 2 \
      2 &4
      end{pmatrix} = nabla ^2 f $$



      Define $phi (t) = f(t, 2t)$. Find $phi ^ prime (0)$ and $phi ^ {prime prime} (0)$.



      Any hints? I know how to do this when $f$ is explicitly defined, but not when just given the gradient/Hessian at a certain point.






      real-analysis







      share|cite|improve this question









      $endgroup$




      Let $F: mathbf{R^2} rightarrow mathbf{R}$ have continuous second order partial derivatives. Assume its gradient is $nabla f(0,0) = (1,2)$ and its Hessian matrix is
      $$begin{pmatrix}
      1 & 2 \
      2 &4
      end{pmatrix} = nabla ^2 f $$



      Define $phi (t) = f(t, 2t)$. Find $phi ^ prime (0)$ and $phi ^ {prime prime} (0)$.



      Any hints? I know how to do this when $f$ is explicitly defined, but not when just given the gradient/Hessian at a certain point.






      real-analysis




      real-analysis






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 17 '18 at 16:01









      hopelessundergradhopelessundergrad

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          1 Answer
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          $begingroup$

          Just aplpy chain rule, nota that $phi=fcirc h$, with $h:mathbb{R}to mathbb{R}^{2}$, $h(t)=(t,2t)$. Then



          begin{equation} phi '(t)=nabla f (h(t)) h'(t) hspace{3cm}(1)
          end{equation}
          then put $t=0$ (h' is a column vector, (1,2) as a column)



          $$ phi '(0)=nabla f (h(0))h'(0) $$



          $$ phi '(0)=nabla f (0,0)h'(0) $$



          $$ phi '(0)=(1,2)(1,2)^{T} $$



          $$ phi '(0)=1+4=5$$



          Then derive again in (1) (which can be writenn as $[(nabla fcirc h)(t)](1,2)^{T}$ to use the chain rule again ) to find $phi''$ taking into account that $h'(t)=(1,2)^{T}$ (constant):



          $$phi ''(t)=[nabla^{2} f (h(t))h'(t) ]left[{begin{array}{c}
          1 \
          2 \
          end{array} } right] $$



          so puting $t=0$, we have:



          $$phi ''(0)=[nabla^{2} f (h(0))h'(0)] left[ {begin{array}{c}
          1 \
          2 \
          end{array} } right] $$



          $$phi ''(0)=[nabla^{2} f (0,0) h'(0)] left[ {begin{array}{c}
          1 \
          2 \
          end{array} } right] $$



          $$phi ''(0)={ left[ {begin{array}{cc}
          1 & 2 \
          2 & 4 \
          end{array} } right] left[ {begin{array}{c}
          1 \
          2 \
          end{array} } right] } left[ {begin{array}{c}
          1 \
          2 \
          end{array} } right] $$



          $$phi ''(0)=left[ {begin{array}{c}
          5 \
          10 \
          end{array} } right] left[{begin{array}{c}
          1 \
          2 \
          end{array} } right] $$



          $$phi''(0)=25.$$






          share|cite|improve this answer











          $endgroup$













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            $begingroup$

            Just aplpy chain rule, nota that $phi=fcirc h$, with $h:mathbb{R}to mathbb{R}^{2}$, $h(t)=(t,2t)$. Then



            begin{equation} phi '(t)=nabla f (h(t)) h'(t) hspace{3cm}(1)
            end{equation}
            then put $t=0$ (h' is a column vector, (1,2) as a column)



            $$ phi '(0)=nabla f (h(0))h'(0) $$



            $$ phi '(0)=nabla f (0,0)h'(0) $$



            $$ phi '(0)=(1,2)(1,2)^{T} $$



            $$ phi '(0)=1+4=5$$



            Then derive again in (1) (which can be writenn as $[(nabla fcirc h)(t)](1,2)^{T}$ to use the chain rule again ) to find $phi''$ taking into account that $h'(t)=(1,2)^{T}$ (constant):



            $$phi ''(t)=[nabla^{2} f (h(t))h'(t) ]left[{begin{array}{c}
            1 \
            2 \
            end{array} } right] $$



            so puting $t=0$, we have:



            $$phi ''(0)=[nabla^{2} f (h(0))h'(0)] left[ {begin{array}{c}
            1 \
            2 \
            end{array} } right] $$



            $$phi ''(0)=[nabla^{2} f (0,0) h'(0)] left[ {begin{array}{c}
            1 \
            2 \
            end{array} } right] $$



            $$phi ''(0)={ left[ {begin{array}{cc}
            1 & 2 \
            2 & 4 \
            end{array} } right] left[ {begin{array}{c}
            1 \
            2 \
            end{array} } right] } left[ {begin{array}{c}
            1 \
            2 \
            end{array} } right] $$



            $$phi ''(0)=left[ {begin{array}{c}
            5 \
            10 \
            end{array} } right] left[{begin{array}{c}
            1 \
            2 \
            end{array} } right] $$



            $$phi''(0)=25.$$






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              Just aplpy chain rule, nota that $phi=fcirc h$, with $h:mathbb{R}to mathbb{R}^{2}$, $h(t)=(t,2t)$. Then



              begin{equation} phi '(t)=nabla f (h(t)) h'(t) hspace{3cm}(1)
              end{equation}
              then put $t=0$ (h' is a column vector, (1,2) as a column)



              $$ phi '(0)=nabla f (h(0))h'(0) $$



              $$ phi '(0)=nabla f (0,0)h'(0) $$



              $$ phi '(0)=(1,2)(1,2)^{T} $$



              $$ phi '(0)=1+4=5$$



              Then derive again in (1) (which can be writenn as $[(nabla fcirc h)(t)](1,2)^{T}$ to use the chain rule again ) to find $phi''$ taking into account that $h'(t)=(1,2)^{T}$ (constant):



              $$phi ''(t)=[nabla^{2} f (h(t))h'(t) ]left[{begin{array}{c}
              1 \
              2 \
              end{array} } right] $$



              so puting $t=0$, we have:



              $$phi ''(0)=[nabla^{2} f (h(0))h'(0)] left[ {begin{array}{c}
              1 \
              2 \
              end{array} } right] $$



              $$phi ''(0)=[nabla^{2} f (0,0) h'(0)] left[ {begin{array}{c}
              1 \
              2 \
              end{array} } right] $$



              $$phi ''(0)={ left[ {begin{array}{cc}
              1 & 2 \
              2 & 4 \
              end{array} } right] left[ {begin{array}{c}
              1 \
              2 \
              end{array} } right] } left[ {begin{array}{c}
              1 \
              2 \
              end{array} } right] $$



              $$phi ''(0)=left[ {begin{array}{c}
              5 \
              10 \
              end{array} } right] left[{begin{array}{c}
              1 \
              2 \
              end{array} } right] $$



              $$phi''(0)=25.$$






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                Just aplpy chain rule, nota that $phi=fcirc h$, with $h:mathbb{R}to mathbb{R}^{2}$, $h(t)=(t,2t)$. Then



                begin{equation} phi '(t)=nabla f (h(t)) h'(t) hspace{3cm}(1)
                end{equation}
                then put $t=0$ (h' is a column vector, (1,2) as a column)



                $$ phi '(0)=nabla f (h(0))h'(0) $$



                $$ phi '(0)=nabla f (0,0)h'(0) $$



                $$ phi '(0)=(1,2)(1,2)^{T} $$



                $$ phi '(0)=1+4=5$$



                Then derive again in (1) (which can be writenn as $[(nabla fcirc h)(t)](1,2)^{T}$ to use the chain rule again ) to find $phi''$ taking into account that $h'(t)=(1,2)^{T}$ (constant):



                $$phi ''(t)=[nabla^{2} f (h(t))h'(t) ]left[{begin{array}{c}
                1 \
                2 \
                end{array} } right] $$



                so puting $t=0$, we have:



                $$phi ''(0)=[nabla^{2} f (h(0))h'(0)] left[ {begin{array}{c}
                1 \
                2 \
                end{array} } right] $$



                $$phi ''(0)=[nabla^{2} f (0,0) h'(0)] left[ {begin{array}{c}
                1 \
                2 \
                end{array} } right] $$



                $$phi ''(0)={ left[ {begin{array}{cc}
                1 & 2 \
                2 & 4 \
                end{array} } right] left[ {begin{array}{c}
                1 \
                2 \
                end{array} } right] } left[ {begin{array}{c}
                1 \
                2 \
                end{array} } right] $$



                $$phi ''(0)=left[ {begin{array}{c}
                5 \
                10 \
                end{array} } right] left[{begin{array}{c}
                1 \
                2 \
                end{array} } right] $$



                $$phi''(0)=25.$$






                share|cite|improve this answer











                $endgroup$



                Just aplpy chain rule, nota that $phi=fcirc h$, with $h:mathbb{R}to mathbb{R}^{2}$, $h(t)=(t,2t)$. Then



                begin{equation} phi '(t)=nabla f (h(t)) h'(t) hspace{3cm}(1)
                end{equation}
                then put $t=0$ (h' is a column vector, (1,2) as a column)



                $$ phi '(0)=nabla f (h(0))h'(0) $$



                $$ phi '(0)=nabla f (0,0)h'(0) $$



                $$ phi '(0)=(1,2)(1,2)^{T} $$



                $$ phi '(0)=1+4=5$$



                Then derive again in (1) (which can be writenn as $[(nabla fcirc h)(t)](1,2)^{T}$ to use the chain rule again ) to find $phi''$ taking into account that $h'(t)=(1,2)^{T}$ (constant):



                $$phi ''(t)=[nabla^{2} f (h(t))h'(t) ]left[{begin{array}{c}
                1 \
                2 \
                end{array} } right] $$



                so puting $t=0$, we have:



                $$phi ''(0)=[nabla^{2} f (h(0))h'(0)] left[ {begin{array}{c}
                1 \
                2 \
                end{array} } right] $$



                $$phi ''(0)=[nabla^{2} f (0,0) h'(0)] left[ {begin{array}{c}
                1 \
                2 \
                end{array} } right] $$



                $$phi ''(0)={ left[ {begin{array}{cc}
                1 & 2 \
                2 & 4 \
                end{array} } right] left[ {begin{array}{c}
                1 \
                2 \
                end{array} } right] } left[ {begin{array}{c}
                1 \
                2 \
                end{array} } right] $$



                $$phi ''(0)=left[ {begin{array}{c}
                5 \
                10 \
                end{array} } right] left[{begin{array}{c}
                1 \
                2 \
                end{array} } right] $$



                $$phi''(0)=25.$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 17 '18 at 22:36

























                answered Dec 17 '18 at 16:08









                Gabriel PalauGabriel Palau

                1106




                1106






























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