Application of consecutive chain rules.












0














Let $C^infty ni f:mathbb{R}^3 to mathbb{R}$ And $xi:mathbb{R}tomathbb{R}$ defined by $xi(t) = f(t,t^2,t^3)$.
I want to find $xi''(t)$.



I defined $g :mathbb{R} to mathbb{R}^3$ as $g(x)=(x,x^2,x^3)$.
Calculation of first order derivative is pretty straigh forward using the chain rule, we have :



$D_xi(t)=D_{fcirc g}(t)=D_f(g(t))D_g(t)=partial_xf(g(t))+2partial_yf(g(t))t+3partial_zf(g(t))t^2$.



However, I have no idea how to differentiate this equation again using the chain rule.



My attempt was :



$frac{d}{dt}(partial_xf(g(t))=D_{partial_xf}(g(t))D_g(t)=partial_{xx}f(g(t))+2partial_{xy}f(g(t))t+3partial_{xz}f(g(t))t^2$



I`m not sure if I applied the chainrule correctly for $partial_xf$.



Moreover, I tried working with the Hessian matrix, I assumed the value of $xi''(t)$ would be $D_g(t)H_f(g(t))D_{D_g(x)}(t)$
Where $D_{D_g(x)}(t)= [0,2,6t]$, and $D_g(t)=[1,2t,3t^2]^T$.



I couldn't prove it however,I'll be glad if someone knows of a way of working with the Hessian in this situation.(I bet it`d be much simpler.)






calculus multivariable-calculus derivatives







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    0














    Let $C^infty ni f:mathbb{R}^3 to mathbb{R}$ And $xi:mathbb{R}tomathbb{R}$ defined by $xi(t) = f(t,t^2,t^3)$.
    I want to find $xi''(t)$.



    I defined $g :mathbb{R} to mathbb{R}^3$ as $g(x)=(x,x^2,x^3)$.
    Calculation of first order derivative is pretty straigh forward using the chain rule, we have :



    $D_xi(t)=D_{fcirc g}(t)=D_f(g(t))D_g(t)=partial_xf(g(t))+2partial_yf(g(t))t+3partial_zf(g(t))t^2$.



    However, I have no idea how to differentiate this equation again using the chain rule.



    My attempt was :



    $frac{d}{dt}(partial_xf(g(t))=D_{partial_xf}(g(t))D_g(t)=partial_{xx}f(g(t))+2partial_{xy}f(g(t))t+3partial_{xz}f(g(t))t^2$



    I`m not sure if I applied the chainrule correctly for $partial_xf$.



    Moreover, I tried working with the Hessian matrix, I assumed the value of $xi''(t)$ would be $D_g(t)H_f(g(t))D_{D_g(x)}(t)$
    Where $D_{D_g(x)}(t)= [0,2,6t]$, and $D_g(t)=[1,2t,3t^2]^T$.



    I couldn't prove it however,I'll be glad if someone knows of a way of working with the Hessian in this situation.(I bet it`d be much simpler.)






    calculus multivariable-calculus derivatives







    share|cite|improve this question

























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      0








      0







      Let $C^infty ni f:mathbb{R}^3 to mathbb{R}$ And $xi:mathbb{R}tomathbb{R}$ defined by $xi(t) = f(t,t^2,t^3)$.
      I want to find $xi''(t)$.



      I defined $g :mathbb{R} to mathbb{R}^3$ as $g(x)=(x,x^2,x^3)$.
      Calculation of first order derivative is pretty straigh forward using the chain rule, we have :



      $D_xi(t)=D_{fcirc g}(t)=D_f(g(t))D_g(t)=partial_xf(g(t))+2partial_yf(g(t))t+3partial_zf(g(t))t^2$.



      However, I have no idea how to differentiate this equation again using the chain rule.



      My attempt was :



      $frac{d}{dt}(partial_xf(g(t))=D_{partial_xf}(g(t))D_g(t)=partial_{xx}f(g(t))+2partial_{xy}f(g(t))t+3partial_{xz}f(g(t))t^2$



      I`m not sure if I applied the chainrule correctly for $partial_xf$.



      Moreover, I tried working with the Hessian matrix, I assumed the value of $xi''(t)$ would be $D_g(t)H_f(g(t))D_{D_g(x)}(t)$
      Where $D_{D_g(x)}(t)= [0,2,6t]$, and $D_g(t)=[1,2t,3t^2]^T$.



      I couldn't prove it however,I'll be glad if someone knows of a way of working with the Hessian in this situation.(I bet it`d be much simpler.)






      calculus multivariable-calculus derivatives







      share|cite|improve this question













      Let $C^infty ni f:mathbb{R}^3 to mathbb{R}$ And $xi:mathbb{R}tomathbb{R}$ defined by $xi(t) = f(t,t^2,t^3)$.
      I want to find $xi''(t)$.



      I defined $g :mathbb{R} to mathbb{R}^3$ as $g(x)=(x,x^2,x^3)$.
      Calculation of first order derivative is pretty straigh forward using the chain rule, we have :



      $D_xi(t)=D_{fcirc g}(t)=D_f(g(t))D_g(t)=partial_xf(g(t))+2partial_yf(g(t))t+3partial_zf(g(t))t^2$.



      However, I have no idea how to differentiate this equation again using the chain rule.



      My attempt was :



      $frac{d}{dt}(partial_xf(g(t))=D_{partial_xf}(g(t))D_g(t)=partial_{xx}f(g(t))+2partial_{xy}f(g(t))t+3partial_{xz}f(g(t))t^2$



      I`m not sure if I applied the chainrule correctly for $partial_xf$.



      Moreover, I tried working with the Hessian matrix, I assumed the value of $xi''(t)$ would be $D_g(t)H_f(g(t))D_{D_g(x)}(t)$
      Where $D_{D_g(x)}(t)= [0,2,6t]$, and $D_g(t)=[1,2t,3t^2]^T$.



      I couldn't prove it however,I'll be glad if someone knows of a way of working with the Hessian in this situation.(I bet it`d be much simpler.)






      calculus multivariable-calculus derivatives




      calculus multivariable-calculus derivatives






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      asked Nov 26 '18 at 18:39









      Sar

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          We have $xi(t)=fbigl({bf r}(t))$, where ${bf r}(t):=(t,t^2,t^3)$. Then
          $$eqalign{xi'(t)&=f_xbigl({bf r}(t)bigr)x'(t)+f_ybigl({bf r}(t)bigr)y'(t)+f_zbigl({bf r}(t))z'(t)cr &=f_xbigl({bf r}(t)bigr)cdot 1+f_ybigl({bf r}(t)bigr)cdot 2t+f_zbigl({bf r}(t))cdot 3t^2 .cr}$$
          For the second derivative we have to use the chain rule again, and also the product rule. I shall omit writing $bigl({bf r}(t)bigr)$ all the time. We obtain
          $$eqalign{xi''(t)&=bigl(f_{xx}cdot 1+f_{xy}cdot 2t+f_{xz}cdot 3t^2bigr)cdot1cr
          & +bigl(f_{yx}cdot1+f_{yy}cdot2t+f_{yz}cdot3t^2bigr)cdot2t+f_ycdot2cr
          & +bigl(f_{zx}cdot1+f_{zy}cdot2t+f_{zz}cdot3t^2bigr)cdot3t^2+f_zcdot 6t ,cr}$$
          which may be somewhat simplified by collecting terms and noting that $f_{xy}=f_{yx}$, etc.






          share|cite|improve this answer





















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            active

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            We have $xi(t)=fbigl({bf r}(t))$, where ${bf r}(t):=(t,t^2,t^3)$. Then
            $$eqalign{xi'(t)&=f_xbigl({bf r}(t)bigr)x'(t)+f_ybigl({bf r}(t)bigr)y'(t)+f_zbigl({bf r}(t))z'(t)cr &=f_xbigl({bf r}(t)bigr)cdot 1+f_ybigl({bf r}(t)bigr)cdot 2t+f_zbigl({bf r}(t))cdot 3t^2 .cr}$$
            For the second derivative we have to use the chain rule again, and also the product rule. I shall omit writing $bigl({bf r}(t)bigr)$ all the time. We obtain
            $$eqalign{xi''(t)&=bigl(f_{xx}cdot 1+f_{xy}cdot 2t+f_{xz}cdot 3t^2bigr)cdot1cr
            & +bigl(f_{yx}cdot1+f_{yy}cdot2t+f_{yz}cdot3t^2bigr)cdot2t+f_ycdot2cr
            & +bigl(f_{zx}cdot1+f_{zy}cdot2t+f_{zz}cdot3t^2bigr)cdot3t^2+f_zcdot 6t ,cr}$$
            which may be somewhat simplified by collecting terms and noting that $f_{xy}=f_{yx}$, etc.






            share|cite|improve this answer


























              1





              +50









              We have $xi(t)=fbigl({bf r}(t))$, where ${bf r}(t):=(t,t^2,t^3)$. Then
              $$eqalign{xi'(t)&=f_xbigl({bf r}(t)bigr)x'(t)+f_ybigl({bf r}(t)bigr)y'(t)+f_zbigl({bf r}(t))z'(t)cr &=f_xbigl({bf r}(t)bigr)cdot 1+f_ybigl({bf r}(t)bigr)cdot 2t+f_zbigl({bf r}(t))cdot 3t^2 .cr}$$
              For the second derivative we have to use the chain rule again, and also the product rule. I shall omit writing $bigl({bf r}(t)bigr)$ all the time. We obtain
              $$eqalign{xi''(t)&=bigl(f_{xx}cdot 1+f_{xy}cdot 2t+f_{xz}cdot 3t^2bigr)cdot1cr
              & +bigl(f_{yx}cdot1+f_{yy}cdot2t+f_{yz}cdot3t^2bigr)cdot2t+f_ycdot2cr
              & +bigl(f_{zx}cdot1+f_{zy}cdot2t+f_{zz}cdot3t^2bigr)cdot3t^2+f_zcdot 6t ,cr}$$
              which may be somewhat simplified by collecting terms and noting that $f_{xy}=f_{yx}$, etc.






              share|cite|improve this answer
























                1





                +50







                1





                +50



                1




                +50




                We have $xi(t)=fbigl({bf r}(t))$, where ${bf r}(t):=(t,t^2,t^3)$. Then
                $$eqalign{xi'(t)&=f_xbigl({bf r}(t)bigr)x'(t)+f_ybigl({bf r}(t)bigr)y'(t)+f_zbigl({bf r}(t))z'(t)cr &=f_xbigl({bf r}(t)bigr)cdot 1+f_ybigl({bf r}(t)bigr)cdot 2t+f_zbigl({bf r}(t))cdot 3t^2 .cr}$$
                For the second derivative we have to use the chain rule again, and also the product rule. I shall omit writing $bigl({bf r}(t)bigr)$ all the time. We obtain
                $$eqalign{xi''(t)&=bigl(f_{xx}cdot 1+f_{xy}cdot 2t+f_{xz}cdot 3t^2bigr)cdot1cr
                & +bigl(f_{yx}cdot1+f_{yy}cdot2t+f_{yz}cdot3t^2bigr)cdot2t+f_ycdot2cr
                & +bigl(f_{zx}cdot1+f_{zy}cdot2t+f_{zz}cdot3t^2bigr)cdot3t^2+f_zcdot 6t ,cr}$$
                which may be somewhat simplified by collecting terms and noting that $f_{xy}=f_{yx}$, etc.






                share|cite|improve this answer












                We have $xi(t)=fbigl({bf r}(t))$, where ${bf r}(t):=(t,t^2,t^3)$. Then
                $$eqalign{xi'(t)&=f_xbigl({bf r}(t)bigr)x'(t)+f_ybigl({bf r}(t)bigr)y'(t)+f_zbigl({bf r}(t))z'(t)cr &=f_xbigl({bf r}(t)bigr)cdot 1+f_ybigl({bf r}(t)bigr)cdot 2t+f_zbigl({bf r}(t))cdot 3t^2 .cr}$$
                For the second derivative we have to use the chain rule again, and also the product rule. I shall omit writing $bigl({bf r}(t)bigr)$ all the time. We obtain
                $$eqalign{xi''(t)&=bigl(f_{xx}cdot 1+f_{xy}cdot 2t+f_{xz}cdot 3t^2bigr)cdot1cr
                & +bigl(f_{yx}cdot1+f_{yy}cdot2t+f_{yz}cdot3t^2bigr)cdot2t+f_ycdot2cr
                & +bigl(f_{zx}cdot1+f_{zy}cdot2t+f_{zz}cdot3t^2bigr)cdot3t^2+f_zcdot 6t ,cr}$$
                which may be somewhat simplified by collecting terms and noting that $f_{xy}=f_{yx}$, etc.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 29 '18 at 19:43









                Christian Blatter

                172k7112325




                172k7112325






























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