Polynomring/Partielle Ableitungen/Variable als Vorfaktor/Aufgabe/Lösung

${}\partial _{1}^{m}(X_{1}\partial ^{\nu })=m\partial _{1}^{m-1}\partial ^{\mu }+X_{1}\partial _{1}^{m}\partial ^{\mu }\,.$

Wir führen Induktion über ${}m$, wobei der Fall ${}m=0$ klar ist. Es ist

${}{\begin{aligned}\partial _{1}^{m}(X_{1}\partial ^{\nu })&=\partial _{1}\partial _{1}^{m-1}(X_{1}\partial ^{\nu })\\&=\partial _{1}{\left((m-1)\partial _{1}^{m-2}\partial ^{\mu }+X_{1}\partial _{1}^{m-1}\partial ^{\mu }\right)}\\&=(m-1)\partial _{1}^{m-1}\partial ^{\mu }+\partial _{1}{\left(X_{1}\partial _{1}^{m-1}\partial ^{\mu }\right)}\\&=(m-1)\partial _{1}^{m-1}\partial ^{\mu }+\partial _{1}^{m-1}\partial ^{\mu }+X_{1}\partial _{1}\partial _{1}^{m-1}\partial ^{\mu }\\&=m\partial _{1}^{m-1}\partial ^{\mu }+X_{1}\partial _{1}^{m}\partial ^{\mu }.\end{aligned}}$

Zur gelösten Aufgabe