Directional derivative of $ f(x)=lVert xrVert^4 $ [on hold]












2












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I'm trying to find the directional derivative of the function $f(mathbf{x})= lVert mathbf{x}rVert^4$ with respect to any vector $mathbf{u}$, where $mathbf{x}$ is a vector in $mathbf{R}^n$.



I tried doing it by definition, but found that it's quite lengthy. Is there any simpler way to do it?



Thanks in advance.






calculus multivariable-calculus derivatives partial-derivative







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put on hold as off-topic by RRL, Cesareo, José Carlos Santos, user91500, Adrian Keister 11 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Cesareo, José Carlos Santos, user91500, Adrian Keister

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  • 1




    $begingroup$
    Do you want the directional derivative at any point? Or at the origin? Because one of these is very simple, but the other I don't see an easy way to simplify. I could be wrong though.
    $endgroup$
    – Joe
    Jan 8 at 18:59






  • 1




    $begingroup$
    Ideally at any point... The one at the origin is simple, yes :)
    $endgroup$
    – SantiMontouliu
    Jan 8 at 19:00






  • 1




    $begingroup$
    Actually I just did it out, and it's not that bad.
    $endgroup$
    – Joe
    Jan 8 at 19:02
















2












$begingroup$


I'm trying to find the directional derivative of the function $f(mathbf{x})= lVert mathbf{x}rVert^4$ with respect to any vector $mathbf{u}$, where $mathbf{x}$ is a vector in $mathbf{R}^n$.



I tried doing it by definition, but found that it's quite lengthy. Is there any simpler way to do it?



Thanks in advance.






calculus multivariable-calculus derivatives partial-derivative







share|cite|improve this question











$endgroup$



put on hold as off-topic by RRL, Cesareo, José Carlos Santos, user91500, Adrian Keister 11 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Cesareo, José Carlos Santos, user91500, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 1




    $begingroup$
    Do you want the directional derivative at any point? Or at the origin? Because one of these is very simple, but the other I don't see an easy way to simplify. I could be wrong though.
    $endgroup$
    – Joe
    Jan 8 at 18:59






  • 1




    $begingroup$
    Ideally at any point... The one at the origin is simple, yes :)
    $endgroup$
    – SantiMontouliu
    Jan 8 at 19:00






  • 1




    $begingroup$
    Actually I just did it out, and it's not that bad.
    $endgroup$
    – Joe
    Jan 8 at 19:02














2












2








2


1



$begingroup$


I'm trying to find the directional derivative of the function $f(mathbf{x})= lVert mathbf{x}rVert^4$ with respect to any vector $mathbf{u}$, where $mathbf{x}$ is a vector in $mathbf{R}^n$.



I tried doing it by definition, but found that it's quite lengthy. Is there any simpler way to do it?



Thanks in advance.






calculus multivariable-calculus derivatives partial-derivative







share|cite|improve this question











$endgroup$




I'm trying to find the directional derivative of the function $f(mathbf{x})= lVert mathbf{x}rVert^4$ with respect to any vector $mathbf{u}$, where $mathbf{x}$ is a vector in $mathbf{R}^n$.



I tried doing it by definition, but found that it's quite lengthy. Is there any simpler way to do it?



Thanks in advance.






calculus multivariable-calculus derivatives partial-derivative




calculus multivariable-calculus derivatives partial-derivative






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share|cite|improve this question













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edited Jan 8 at 18:58









user1337

3298




3298










asked Jan 8 at 18:57









SantiMontouliuSantiMontouliu

628




628




put on hold as off-topic by RRL, Cesareo, José Carlos Santos, user91500, Adrian Keister 11 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Cesareo, José Carlos Santos, user91500, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by RRL, Cesareo, José Carlos Santos, user91500, Adrian Keister 11 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Cesareo, José Carlos Santos, user91500, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    $begingroup$
    Do you want the directional derivative at any point? Or at the origin? Because one of these is very simple, but the other I don't see an easy way to simplify. I could be wrong though.
    $endgroup$
    – Joe
    Jan 8 at 18:59






  • 1




    $begingroup$
    Ideally at any point... The one at the origin is simple, yes :)
    $endgroup$
    – SantiMontouliu
    Jan 8 at 19:00






  • 1




    $begingroup$
    Actually I just did it out, and it's not that bad.
    $endgroup$
    – Joe
    Jan 8 at 19:02














  • 1




    $begingroup$
    Do you want the directional derivative at any point? Or at the origin? Because one of these is very simple, but the other I don't see an easy way to simplify. I could be wrong though.
    $endgroup$
    – Joe
    Jan 8 at 18:59






  • 1




    $begingroup$
    Ideally at any point... The one at the origin is simple, yes :)
    $endgroup$
    – SantiMontouliu
    Jan 8 at 19:00






  • 1




    $begingroup$
    Actually I just did it out, and it's not that bad.
    $endgroup$
    – Joe
    Jan 8 at 19:02








1




1




$begingroup$
Do you want the directional derivative at any point? Or at the origin? Because one of these is very simple, but the other I don't see an easy way to simplify. I could be wrong though.
$endgroup$
– Joe
Jan 8 at 18:59




$begingroup$
Do you want the directional derivative at any point? Or at the origin? Because one of these is very simple, but the other I don't see an easy way to simplify. I could be wrong though.
$endgroup$
– Joe
Jan 8 at 18:59




1




1




$begingroup$
Ideally at any point... The one at the origin is simple, yes :)
$endgroup$
– SantiMontouliu
Jan 8 at 19:00




$begingroup$
Ideally at any point... The one at the origin is simple, yes :)
$endgroup$
– SantiMontouliu
Jan 8 at 19:00




1




1




$begingroup$
Actually I just did it out, and it's not that bad.
$endgroup$
– Joe
Jan 8 at 19:02




$begingroup$
Actually I just did it out, and it's not that bad.
$endgroup$
– Joe
Jan 8 at 19:02










3 Answers
3






active

oldest

votes


















3












$begingroup$

This function is easy enough to visualize for the purpose of computing its gradient. The gradient is parallel to $vec{x}$, and has magnitude $4lVert vec{x}rVert^3$. So $$nabla f=4lVert vec{x}rVert^2vec{x}$$
Now use the fact that $$D_{vec{u}}=nabla fcdotvec{u}$$ (Assuming $vec{u}$ is a unit vector. Otherwise replace $vec{u}$ with $vec{u}/lVertvec{u}rVert$.)






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  • 1




    $begingroup$
    Thanks a lot for your answer, Alex! By the way, I'm a huge fan of your triangle notation (math.stackexchange.com/questions/30046/…) :)
    $endgroup$
    – SantiMontouliu
    Jan 8 at 19:13










  • $begingroup$
    Every once in a while I have an interesting idea. Thanks for letting me know :)
    $endgroup$
    – alex.jordan
    Jan 8 at 19:20












  • $begingroup$
    It was so cool having you answer one of my questions :D
    $endgroup$
    – SantiMontouliu
    Jan 8 at 20:47



















5












$begingroup$

Since $$frac{partial f}{partial x_i}(mathbf x) = frac{partial}{partial x_i} (x_1^2 + cdots + x_n^2)^2 = 2(x_1^2 + cdots + x_n^2)cdot 2x_i = 4 |mathbf x|^2 x_i$$
you have
$$nabla f(mathbf x) = 4 | mathbf x|^2 mathbf x$$
so that for any unit vector $mathbf u$
$$frac{partial f}{partial mathbf u}(mathbf x) = 4 | mathbf x|^2 (mathbf x cdot mathbf u).$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your answer @Umberto, it was very clear!
    $endgroup$
    – SantiMontouliu
    Jan 8 at 19:12



















0












$begingroup$

$f(mathbf x) = Vert mathbf x Vert^4 = (Vert mathbf x Vert^2)^2 = (langle mathbf x, mathbf x rangle)^2; tag 1$



then



$nabla_{mathbf u}f(mathbf x) = nabla_u (langle mathbf x, mathbf x rangle)^2 = 2langle mathbf x, mathbf x rangle nabla_{mathbf u} langle mathbf x, mathbf x rangle = 2langle mathbf x, mathbf x rangle (2langle mathbf x, nabla_{mathbf u} mathbf x rangle) = 4Vert mathbf x Vert^2 langle mathbf x, nabla_{mathbf u} mathbf x rangle; tag 2$



also,



$nabla_{mathbf u} mathbf x = (nabla_{mathbf u} x_1, nabla_{mathbf u} x_2, ldots, nabla_{mathbf u} x_n), tag 3$



and



$nabla_{mathbf u} x_i = left (displaystyle sum_1^n u_j dfrac{partial}{partial x_j} right ) x_i = displaystyle sum_1^n u_j dfrac{partial x_i}{partial x_j}= sum_1^n u_j delta_{ij} = u_i, tag 4$



whence



$nabla_{mathbf u} mathbf x = (u_1, u_2, ldots, u_n), tag 5$



and thus



$langle mathbf x, nabla_{mathbf u} mathbf x rangle = langle mathbf x, mathbf u rangle; tag 6$



therefore (2) becomes



$nabla_{mathbf u}f(mathbf x) = 4Vert mathbf x Vert^2 langle mathbf x, mathbf u rangle. tag 7$



We thus clearly have



$nabla_{mathbf u}f(0) = 0 tag 8$



as well.






share|cite|improve this answer









$endgroup$




















    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    This function is easy enough to visualize for the purpose of computing its gradient. The gradient is parallel to $vec{x}$, and has magnitude $4lVert vec{x}rVert^3$. So $$nabla f=4lVert vec{x}rVert^2vec{x}$$
    Now use the fact that $$D_{vec{u}}=nabla fcdotvec{u}$$ (Assuming $vec{u}$ is a unit vector. Otherwise replace $vec{u}$ with $vec{u}/lVertvec{u}rVert$.)






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Thanks a lot for your answer, Alex! By the way, I'm a huge fan of your triangle notation (math.stackexchange.com/questions/30046/…) :)
      $endgroup$
      – SantiMontouliu
      Jan 8 at 19:13










    • $begingroup$
      Every once in a while I have an interesting idea. Thanks for letting me know :)
      $endgroup$
      – alex.jordan
      Jan 8 at 19:20












    • $begingroup$
      It was so cool having you answer one of my questions :D
      $endgroup$
      – SantiMontouliu
      Jan 8 at 20:47
















    3












    $begingroup$

    This function is easy enough to visualize for the purpose of computing its gradient. The gradient is parallel to $vec{x}$, and has magnitude $4lVert vec{x}rVert^3$. So $$nabla f=4lVert vec{x}rVert^2vec{x}$$
    Now use the fact that $$D_{vec{u}}=nabla fcdotvec{u}$$ (Assuming $vec{u}$ is a unit vector. Otherwise replace $vec{u}$ with $vec{u}/lVertvec{u}rVert$.)






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Thanks a lot for your answer, Alex! By the way, I'm a huge fan of your triangle notation (math.stackexchange.com/questions/30046/…) :)
      $endgroup$
      – SantiMontouliu
      Jan 8 at 19:13










    • $begingroup$
      Every once in a while I have an interesting idea. Thanks for letting me know :)
      $endgroup$
      – alex.jordan
      Jan 8 at 19:20












    • $begingroup$
      It was so cool having you answer one of my questions :D
      $endgroup$
      – SantiMontouliu
      Jan 8 at 20:47














    3












    3








    3





    $begingroup$

    This function is easy enough to visualize for the purpose of computing its gradient. The gradient is parallel to $vec{x}$, and has magnitude $4lVert vec{x}rVert^3$. So $$nabla f=4lVert vec{x}rVert^2vec{x}$$
    Now use the fact that $$D_{vec{u}}=nabla fcdotvec{u}$$ (Assuming $vec{u}$ is a unit vector. Otherwise replace $vec{u}$ with $vec{u}/lVertvec{u}rVert$.)






    share|cite|improve this answer









    $endgroup$



    This function is easy enough to visualize for the purpose of computing its gradient. The gradient is parallel to $vec{x}$, and has magnitude $4lVert vec{x}rVert^3$. So $$nabla f=4lVert vec{x}rVert^2vec{x}$$
    Now use the fact that $$D_{vec{u}}=nabla fcdotvec{u}$$ (Assuming $vec{u}$ is a unit vector. Otherwise replace $vec{u}$ with $vec{u}/lVertvec{u}rVert$.)







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 8 at 19:04









    alex.jordanalex.jordan

    39k560120




    39k560120








    • 1




      $begingroup$
      Thanks a lot for your answer, Alex! By the way, I'm a huge fan of your triangle notation (math.stackexchange.com/questions/30046/…) :)
      $endgroup$
      – SantiMontouliu
      Jan 8 at 19:13










    • $begingroup$
      Every once in a while I have an interesting idea. Thanks for letting me know :)
      $endgroup$
      – alex.jordan
      Jan 8 at 19:20












    • $begingroup$
      It was so cool having you answer one of my questions :D
      $endgroup$
      – SantiMontouliu
      Jan 8 at 20:47














    • 1




      $begingroup$
      Thanks a lot for your answer, Alex! By the way, I'm a huge fan of your triangle notation (math.stackexchange.com/questions/30046/…) :)
      $endgroup$
      – SantiMontouliu
      Jan 8 at 19:13










    • $begingroup$
      Every once in a while I have an interesting idea. Thanks for letting me know :)
      $endgroup$
      – alex.jordan
      Jan 8 at 19:20












    • $begingroup$
      It was so cool having you answer one of my questions :D
      $endgroup$
      – SantiMontouliu
      Jan 8 at 20:47








    1




    1




    $begingroup$
    Thanks a lot for your answer, Alex! By the way, I'm a huge fan of your triangle notation (math.stackexchange.com/questions/30046/…) :)
    $endgroup$
    – SantiMontouliu
    Jan 8 at 19:13




    $begingroup$
    Thanks a lot for your answer, Alex! By the way, I'm a huge fan of your triangle notation (math.stackexchange.com/questions/30046/…) :)
    $endgroup$
    – SantiMontouliu
    Jan 8 at 19:13












    $begingroup$
    Every once in a while I have an interesting idea. Thanks for letting me know :)
    $endgroup$
    – alex.jordan
    Jan 8 at 19:20






    $begingroup$
    Every once in a while I have an interesting idea. Thanks for letting me know :)
    $endgroup$
    – alex.jordan
    Jan 8 at 19:20














    $begingroup$
    It was so cool having you answer one of my questions :D
    $endgroup$
    – SantiMontouliu
    Jan 8 at 20:47




    $begingroup$
    It was so cool having you answer one of my questions :D
    $endgroup$
    – SantiMontouliu
    Jan 8 at 20:47











    5












    $begingroup$

    Since $$frac{partial f}{partial x_i}(mathbf x) = frac{partial}{partial x_i} (x_1^2 + cdots + x_n^2)^2 = 2(x_1^2 + cdots + x_n^2)cdot 2x_i = 4 |mathbf x|^2 x_i$$
    you have
    $$nabla f(mathbf x) = 4 | mathbf x|^2 mathbf x$$
    so that for any unit vector $mathbf u$
    $$frac{partial f}{partial mathbf u}(mathbf x) = 4 | mathbf x|^2 (mathbf x cdot mathbf u).$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks for your answer @Umberto, it was very clear!
      $endgroup$
      – SantiMontouliu
      Jan 8 at 19:12
















    5












    $begingroup$

    Since $$frac{partial f}{partial x_i}(mathbf x) = frac{partial}{partial x_i} (x_1^2 + cdots + x_n^2)^2 = 2(x_1^2 + cdots + x_n^2)cdot 2x_i = 4 |mathbf x|^2 x_i$$
    you have
    $$nabla f(mathbf x) = 4 | mathbf x|^2 mathbf x$$
    so that for any unit vector $mathbf u$
    $$frac{partial f}{partial mathbf u}(mathbf x) = 4 | mathbf x|^2 (mathbf x cdot mathbf u).$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks for your answer @Umberto, it was very clear!
      $endgroup$
      – SantiMontouliu
      Jan 8 at 19:12














    5












    5








    5





    $begingroup$

    Since $$frac{partial f}{partial x_i}(mathbf x) = frac{partial}{partial x_i} (x_1^2 + cdots + x_n^2)^2 = 2(x_1^2 + cdots + x_n^2)cdot 2x_i = 4 |mathbf x|^2 x_i$$
    you have
    $$nabla f(mathbf x) = 4 | mathbf x|^2 mathbf x$$
    so that for any unit vector $mathbf u$
    $$frac{partial f}{partial mathbf u}(mathbf x) = 4 | mathbf x|^2 (mathbf x cdot mathbf u).$$






    share|cite|improve this answer









    $endgroup$



    Since $$frac{partial f}{partial x_i}(mathbf x) = frac{partial}{partial x_i} (x_1^2 + cdots + x_n^2)^2 = 2(x_1^2 + cdots + x_n^2)cdot 2x_i = 4 |mathbf x|^2 x_i$$
    you have
    $$nabla f(mathbf x) = 4 | mathbf x|^2 mathbf x$$
    so that for any unit vector $mathbf u$
    $$frac{partial f}{partial mathbf u}(mathbf x) = 4 | mathbf x|^2 (mathbf x cdot mathbf u).$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 8 at 19:04









    Umberto P.Umberto P.

    38.8k13064




    38.8k13064












    • $begingroup$
      Thanks for your answer @Umberto, it was very clear!
      $endgroup$
      – SantiMontouliu
      Jan 8 at 19:12


















    • $begingroup$
      Thanks for your answer @Umberto, it was very clear!
      $endgroup$
      – SantiMontouliu
      Jan 8 at 19:12
















    $begingroup$
    Thanks for your answer @Umberto, it was very clear!
    $endgroup$
    – SantiMontouliu
    Jan 8 at 19:12




    $begingroup$
    Thanks for your answer @Umberto, it was very clear!
    $endgroup$
    – SantiMontouliu
    Jan 8 at 19:12











    0












    $begingroup$

    $f(mathbf x) = Vert mathbf x Vert^4 = (Vert mathbf x Vert^2)^2 = (langle mathbf x, mathbf x rangle)^2; tag 1$



    then



    $nabla_{mathbf u}f(mathbf x) = nabla_u (langle mathbf x, mathbf x rangle)^2 = 2langle mathbf x, mathbf x rangle nabla_{mathbf u} langle mathbf x, mathbf x rangle = 2langle mathbf x, mathbf x rangle (2langle mathbf x, nabla_{mathbf u} mathbf x rangle) = 4Vert mathbf x Vert^2 langle mathbf x, nabla_{mathbf u} mathbf x rangle; tag 2$



    also,



    $nabla_{mathbf u} mathbf x = (nabla_{mathbf u} x_1, nabla_{mathbf u} x_2, ldots, nabla_{mathbf u} x_n), tag 3$



    and



    $nabla_{mathbf u} x_i = left (displaystyle sum_1^n u_j dfrac{partial}{partial x_j} right ) x_i = displaystyle sum_1^n u_j dfrac{partial x_i}{partial x_j}= sum_1^n u_j delta_{ij} = u_i, tag 4$



    whence



    $nabla_{mathbf u} mathbf x = (u_1, u_2, ldots, u_n), tag 5$



    and thus



    $langle mathbf x, nabla_{mathbf u} mathbf x rangle = langle mathbf x, mathbf u rangle; tag 6$



    therefore (2) becomes



    $nabla_{mathbf u}f(mathbf x) = 4Vert mathbf x Vert^2 langle mathbf x, mathbf u rangle. tag 7$



    We thus clearly have



    $nabla_{mathbf u}f(0) = 0 tag 8$



    as well.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      $f(mathbf x) = Vert mathbf x Vert^4 = (Vert mathbf x Vert^2)^2 = (langle mathbf x, mathbf x rangle)^2; tag 1$



      then



      $nabla_{mathbf u}f(mathbf x) = nabla_u (langle mathbf x, mathbf x rangle)^2 = 2langle mathbf x, mathbf x rangle nabla_{mathbf u} langle mathbf x, mathbf x rangle = 2langle mathbf x, mathbf x rangle (2langle mathbf x, nabla_{mathbf u} mathbf x rangle) = 4Vert mathbf x Vert^2 langle mathbf x, nabla_{mathbf u} mathbf x rangle; tag 2$



      also,



      $nabla_{mathbf u} mathbf x = (nabla_{mathbf u} x_1, nabla_{mathbf u} x_2, ldots, nabla_{mathbf u} x_n), tag 3$



      and



      $nabla_{mathbf u} x_i = left (displaystyle sum_1^n u_j dfrac{partial}{partial x_j} right ) x_i = displaystyle sum_1^n u_j dfrac{partial x_i}{partial x_j}= sum_1^n u_j delta_{ij} = u_i, tag 4$



      whence



      $nabla_{mathbf u} mathbf x = (u_1, u_2, ldots, u_n), tag 5$



      and thus



      $langle mathbf x, nabla_{mathbf u} mathbf x rangle = langle mathbf x, mathbf u rangle; tag 6$



      therefore (2) becomes



      $nabla_{mathbf u}f(mathbf x) = 4Vert mathbf x Vert^2 langle mathbf x, mathbf u rangle. tag 7$



      We thus clearly have



      $nabla_{mathbf u}f(0) = 0 tag 8$



      as well.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        $f(mathbf x) = Vert mathbf x Vert^4 = (Vert mathbf x Vert^2)^2 = (langle mathbf x, mathbf x rangle)^2; tag 1$



        then



        $nabla_{mathbf u}f(mathbf x) = nabla_u (langle mathbf x, mathbf x rangle)^2 = 2langle mathbf x, mathbf x rangle nabla_{mathbf u} langle mathbf x, mathbf x rangle = 2langle mathbf x, mathbf x rangle (2langle mathbf x, nabla_{mathbf u} mathbf x rangle) = 4Vert mathbf x Vert^2 langle mathbf x, nabla_{mathbf u} mathbf x rangle; tag 2$



        also,



        $nabla_{mathbf u} mathbf x = (nabla_{mathbf u} x_1, nabla_{mathbf u} x_2, ldots, nabla_{mathbf u} x_n), tag 3$



        and



        $nabla_{mathbf u} x_i = left (displaystyle sum_1^n u_j dfrac{partial}{partial x_j} right ) x_i = displaystyle sum_1^n u_j dfrac{partial x_i}{partial x_j}= sum_1^n u_j delta_{ij} = u_i, tag 4$



        whence



        $nabla_{mathbf u} mathbf x = (u_1, u_2, ldots, u_n), tag 5$



        and thus



        $langle mathbf x, nabla_{mathbf u} mathbf x rangle = langle mathbf x, mathbf u rangle; tag 6$



        therefore (2) becomes



        $nabla_{mathbf u}f(mathbf x) = 4Vert mathbf x Vert^2 langle mathbf x, mathbf u rangle. tag 7$



        We thus clearly have



        $nabla_{mathbf u}f(0) = 0 tag 8$



        as well.






        share|cite|improve this answer









        $endgroup$



        $f(mathbf x) = Vert mathbf x Vert^4 = (Vert mathbf x Vert^2)^2 = (langle mathbf x, mathbf x rangle)^2; tag 1$



        then



        $nabla_{mathbf u}f(mathbf x) = nabla_u (langle mathbf x, mathbf x rangle)^2 = 2langle mathbf x, mathbf x rangle nabla_{mathbf u} langle mathbf x, mathbf x rangle = 2langle mathbf x, mathbf x rangle (2langle mathbf x, nabla_{mathbf u} mathbf x rangle) = 4Vert mathbf x Vert^2 langle mathbf x, nabla_{mathbf u} mathbf x rangle; tag 2$



        also,



        $nabla_{mathbf u} mathbf x = (nabla_{mathbf u} x_1, nabla_{mathbf u} x_2, ldots, nabla_{mathbf u} x_n), tag 3$



        and



        $nabla_{mathbf u} x_i = left (displaystyle sum_1^n u_j dfrac{partial}{partial x_j} right ) x_i = displaystyle sum_1^n u_j dfrac{partial x_i}{partial x_j}= sum_1^n u_j delta_{ij} = u_i, tag 4$



        whence



        $nabla_{mathbf u} mathbf x = (u_1, u_2, ldots, u_n), tag 5$



        and thus



        $langle mathbf x, nabla_{mathbf u} mathbf x rangle = langle mathbf x, mathbf u rangle; tag 6$



        therefore (2) becomes



        $nabla_{mathbf u}f(mathbf x) = 4Vert mathbf x Vert^2 langle mathbf x, mathbf u rangle. tag 7$



        We thus clearly have



        $nabla_{mathbf u}f(0) = 0 tag 8$



        as well.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 19:34









        Robert LewisRobert Lewis

        44.6k22964




        44.6k22964















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