Derivative of $x^{beta}$












1














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










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
















1














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
























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










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














1












1








1







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















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













share|cite|improve this question




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edited Dec 4 '18 at 11:10









Harry49

5,99121031




5,99121031










asked Dec 3 '18 at 20:06









aymen

304




304












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










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


















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










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
















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




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












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




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










1 Answer
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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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    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


























      1














      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
























        1












        1








        1






        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












        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 Aguirre

        67.6k24094




        67.6k24094






























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