Derivative of $x^{beta}$












1












$begingroup$


Let $xinmathbb{R}^d$, and $alpha,betainmathbb{N}^d$ such that $|alpha|inleft{1,2right}$ $|beta|ge 3$



What is the result of $$partial_x^{alpha}x^{beta}$$?



Thanks






real-analysis partial-derivative







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  • $begingroup$
    So $alpha$ and $beta$ are vectors whose components can only be natural numbers (in addition to the other restrictions)?
    $endgroup$
    – Chase Ryan Taylor
    Dec 3 '18 at 20:21










  • $begingroup$
    yes that is it exactly
    $endgroup$
    – aymen
    Dec 3 '18 at 20:21
















1












$begingroup$


Let $xinmathbb{R}^d$, and $alpha,betainmathbb{N}^d$ such that $|alpha|inleft{1,2right}$ $|beta|ge 3$



What is the result of $$partial_x^{alpha}x^{beta}$$?



Thanks






real-analysis partial-derivative







share|cite|improve this question











$endgroup$












  • $begingroup$
    So $alpha$ and $beta$ are vectors whose components can only be natural numbers (in addition to the other restrictions)?
    $endgroup$
    – Chase Ryan Taylor
    Dec 3 '18 at 20:21










  • $begingroup$
    yes that is it exactly
    $endgroup$
    – aymen
    Dec 3 '18 at 20:21














1












1








1





$begingroup$


Let $xinmathbb{R}^d$, and $alpha,betainmathbb{N}^d$ such that $|alpha|inleft{1,2right}$ $|beta|ge 3$



What is the result of $$partial_x^{alpha}x^{beta}$$?



Thanks






real-analysis partial-derivative







share|cite|improve this question











$endgroup$




Let $xinmathbb{R}^d$, and $alpha,betainmathbb{N}^d$ such that $|alpha|inleft{1,2right}$ $|beta|ge 3$



What is the result of $$partial_x^{alpha}x^{beta}$$?



Thanks






real-analysis partial-derivative




real-analysis partial-derivative






share|cite|improve this question















share|cite|improve this question













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edited Dec 4 '18 at 11:10









Harry49

6,17331132




6,17331132










asked Dec 3 '18 at 20:06









aymenaymen

304




304












  • $begingroup$
    So $alpha$ and $beta$ are vectors whose components can only be natural numbers (in addition to the other restrictions)?
    $endgroup$
    – Chase Ryan Taylor
    Dec 3 '18 at 20:21










  • $begingroup$
    yes that is it exactly
    $endgroup$
    – aymen
    Dec 3 '18 at 20:21


















  • $begingroup$
    So $alpha$ and $beta$ are vectors whose components can only be natural numbers (in addition to the other restrictions)?
    $endgroup$
    – Chase Ryan Taylor
    Dec 3 '18 at 20:21










  • $begingroup$
    yes that is it exactly
    $endgroup$
    – aymen
    Dec 3 '18 at 20:21
















$begingroup$
So $alpha$ and $beta$ are vectors whose components can only be natural numbers (in addition to the other restrictions)?
$endgroup$
– Chase Ryan Taylor
Dec 3 '18 at 20:21




$begingroup$
So $alpha$ and $beta$ are vectors whose components can only be natural numbers (in addition to the other restrictions)?
$endgroup$
– Chase Ryan Taylor
Dec 3 '18 at 20:21












$begingroup$
yes that is it exactly
$endgroup$
– aymen
Dec 3 '18 at 20:21




$begingroup$
yes that is it exactly
$endgroup$
– aymen
Dec 3 '18 at 20:21










1 Answer
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$begingroup$

Let $alpha=(alpha_1,dots,alpha_n)$ and $beta=(beta_1,dots,beta_n)$. If $alphalebeta$, that is, $alpha_klebeta_k$, $1le kle n$,Then
begin{align}
partial_x^alpha x^beta&=prod_{k=1}^nfrac{partial^{alpha_k} x_k^{beta_k}}{partial x_k^{alpha_k}}\
&=prod_{k=1}^nbeta_k(beta_k-1)dots(beta_k-alpha_k+1)x_k^{beta_k-alpha_k}\
&=Bigl(prod_{k=1}^nfrac{beta_k!}{(beta_k-alpha_k)!}Bigr)x^{beta-alpha}\
&=frac{beta!}{(beta-alpha)!},x^{beta-alpha}.
end{align}






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    $begingroup$

    Let $alpha=(alpha_1,dots,alpha_n)$ and $beta=(beta_1,dots,beta_n)$. If $alphalebeta$, that is, $alpha_klebeta_k$, $1le kle n$,Then
    begin{align}
    partial_x^alpha x^beta&=prod_{k=1}^nfrac{partial^{alpha_k} x_k^{beta_k}}{partial x_k^{alpha_k}}\
    &=prod_{k=1}^nbeta_k(beta_k-1)dots(beta_k-alpha_k+1)x_k^{beta_k-alpha_k}\
    &=Bigl(prod_{k=1}^nfrac{beta_k!}{(beta_k-alpha_k)!}Bigr)x^{beta-alpha}\
    &=frac{beta!}{(beta-alpha)!},x^{beta-alpha}.
    end{align}






    share|cite|improve this answer









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      1












      $begingroup$

      Let $alpha=(alpha_1,dots,alpha_n)$ and $beta=(beta_1,dots,beta_n)$. If $alphalebeta$, that is, $alpha_klebeta_k$, $1le kle n$,Then
      begin{align}
      partial_x^alpha x^beta&=prod_{k=1}^nfrac{partial^{alpha_k} x_k^{beta_k}}{partial x_k^{alpha_k}}\
      &=prod_{k=1}^nbeta_k(beta_k-1)dots(beta_k-alpha_k+1)x_k^{beta_k-alpha_k}\
      &=Bigl(prod_{k=1}^nfrac{beta_k!}{(beta_k-alpha_k)!}Bigr)x^{beta-alpha}\
      &=frac{beta!}{(beta-alpha)!},x^{beta-alpha}.
      end{align}






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Let $alpha=(alpha_1,dots,alpha_n)$ and $beta=(beta_1,dots,beta_n)$. If $alphalebeta$, that is, $alpha_klebeta_k$, $1le kle n$,Then
        begin{align}
        partial_x^alpha x^beta&=prod_{k=1}^nfrac{partial^{alpha_k} x_k^{beta_k}}{partial x_k^{alpha_k}}\
        &=prod_{k=1}^nbeta_k(beta_k-1)dots(beta_k-alpha_k+1)x_k^{beta_k-alpha_k}\
        &=Bigl(prod_{k=1}^nfrac{beta_k!}{(beta_k-alpha_k)!}Bigr)x^{beta-alpha}\
        &=frac{beta!}{(beta-alpha)!},x^{beta-alpha}.
        end{align}






        share|cite|improve this answer









        $endgroup$



        Let $alpha=(alpha_1,dots,alpha_n)$ and $beta=(beta_1,dots,beta_n)$. If $alphalebeta$, that is, $alpha_klebeta_k$, $1le kle n$,Then
        begin{align}
        partial_x^alpha x^beta&=prod_{k=1}^nfrac{partial^{alpha_k} x_k^{beta_k}}{partial x_k^{alpha_k}}\
        &=prod_{k=1}^nbeta_k(beta_k-1)dots(beta_k-alpha_k+1)x_k^{beta_k-alpha_k}\
        &=Bigl(prod_{k=1}^nfrac{beta_k!}{(beta_k-alpha_k)!}Bigr)x^{beta-alpha}\
        &=frac{beta!}{(beta-alpha)!},x^{beta-alpha}.
        end{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 '18 at 15:40









        Julián AguirreJulián Aguirre

        68.4k24094




        68.4k24094






























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