Directional derivative and manipulation of nonlinear functions












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$begingroup$


I have a question about a few manipulations when computing a directional derivative at the point $(0,0)$.



We have $f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h}$ where $f(x,y) = (xy)^frac{1}{3}$ and $u = (u_1, u_2)$. This is just the definition of a directional derivative.



Then we expand this to: $f'(0,u) = limlimits_{h to 0} frac{(h u_1h u_2)^frac{1}{3}}{h}$



I don't understand the steps of this fully. Are the precise steps:



$f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h} = limlimits_{h to 0} frac{f(h(u_1,u_2))}{h}=limlimits_{h to 0} frac{f((hu_1,hu_2))}{h}$



This seems to make sense to me but I just want to make sure I am allowed to do these operations inside the function (in particular, taking the scalar $h$ inside in the last step).



Thanks for your help.






calculus derivatives







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    -1












    $begingroup$


    I have a question about a few manipulations when computing a directional derivative at the point $(0,0)$.



    We have $f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h}$ where $f(x,y) = (xy)^frac{1}{3}$ and $u = (u_1, u_2)$. This is just the definition of a directional derivative.



    Then we expand this to: $f'(0,u) = limlimits_{h to 0} frac{(h u_1h u_2)^frac{1}{3}}{h}$



    I don't understand the steps of this fully. Are the precise steps:



    $f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h} = limlimits_{h to 0} frac{f(h(u_1,u_2))}{h}=limlimits_{h to 0} frac{f((hu_1,hu_2))}{h}$



    This seems to make sense to me but I just want to make sure I am allowed to do these operations inside the function (in particular, taking the scalar $h$ inside in the last step).



    Thanks for your help.






    calculus derivatives







    share|cite|improve this question









    $endgroup$















      -1












      -1








      -1





      $begingroup$


      I have a question about a few manipulations when computing a directional derivative at the point $(0,0)$.



      We have $f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h}$ where $f(x,y) = (xy)^frac{1}{3}$ and $u = (u_1, u_2)$. This is just the definition of a directional derivative.



      Then we expand this to: $f'(0,u) = limlimits_{h to 0} frac{(h u_1h u_2)^frac{1}{3}}{h}$



      I don't understand the steps of this fully. Are the precise steps:



      $f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h} = limlimits_{h to 0} frac{f(h(u_1,u_2))}{h}=limlimits_{h to 0} frac{f((hu_1,hu_2))}{h}$



      This seems to make sense to me but I just want to make sure I am allowed to do these operations inside the function (in particular, taking the scalar $h$ inside in the last step).



      Thanks for your help.






      calculus derivatives







      share|cite|improve this question









      $endgroup$




      I have a question about a few manipulations when computing a directional derivative at the point $(0,0)$.



      We have $f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h}$ where $f(x,y) = (xy)^frac{1}{3}$ and $u = (u_1, u_2)$. This is just the definition of a directional derivative.



      Then we expand this to: $f'(0,u) = limlimits_{h to 0} frac{(h u_1h u_2)^frac{1}{3}}{h}$



      I don't understand the steps of this fully. Are the precise steps:



      $f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h} = limlimits_{h to 0} frac{f(h(u_1,u_2))}{h}=limlimits_{h to 0} frac{f((hu_1,hu_2))}{h}$



      This seems to make sense to me but I just want to make sure I am allowed to do these operations inside the function (in particular, taking the scalar $h$ inside in the last step).



      Thanks for your help.






      calculus derivatives




      calculus derivatives






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      asked Dec 6 '18 at 0:06









      ChristianChristian

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          $begingroup$

          Yes we have



          $$f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h} = limlimits_{h to 0} frac{f(h(u_1,u_2))}{h}=limlimits_{h to 0} frac{f((hu_1,hu_2))}{h}=limlimits_{h to 0} frac{h^frac23(u_1u_2)^frac13}{h}$$



          and for $u_1,u_2 neq 0$ the directional derivative doesn't exist.






          share|cite|improve this answer









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            $begingroup$

            Yes we have



            $$f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h} = limlimits_{h to 0} frac{f(h(u_1,u_2))}{h}=limlimits_{h to 0} frac{f((hu_1,hu_2))}{h}=limlimits_{h to 0} frac{h^frac23(u_1u_2)^frac13}{h}$$



            and for $u_1,u_2 neq 0$ the directional derivative doesn't exist.






            share|cite|improve this answer









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              0












              $begingroup$

              Yes we have



              $$f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h} = limlimits_{h to 0} frac{f(h(u_1,u_2))}{h}=limlimits_{h to 0} frac{f((hu_1,hu_2))}{h}=limlimits_{h to 0} frac{h^frac23(u_1u_2)^frac13}{h}$$



              and for $u_1,u_2 neq 0$ the directional derivative doesn't exist.






              share|cite|improve this answer









              $endgroup$
















                0












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                $begingroup$

                Yes we have



                $$f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h} = limlimits_{h to 0} frac{f(h(u_1,u_2))}{h}=limlimits_{h to 0} frac{f((hu_1,hu_2))}{h}=limlimits_{h to 0} frac{h^frac23(u_1u_2)^frac13}{h}$$



                and for $u_1,u_2 neq 0$ the directional derivative doesn't exist.






                share|cite|improve this answer









                $endgroup$



                Yes we have



                $$f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h} = limlimits_{h to 0} frac{f(h(u_1,u_2))}{h}=limlimits_{h to 0} frac{f((hu_1,hu_2))}{h}=limlimits_{h to 0} frac{h^frac23(u_1u_2)^frac13}{h}$$



                and for $u_1,u_2 neq 0$ the directional derivative doesn't exist.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 6 '18 at 0:10









                gimusigimusi

                92.8k84494




                92.8k84494






























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