Identität/Skalare Funktion/Ableitung/Aufgabe/Lösung

1. In den Koordinaten ${\displaystyle {}x_{1},\ldots ,x_{n}}$ wird die Abbildung durch
${\displaystyle {\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}\mapsto g(x_{1},\ldots ,x_{n}){\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}={\begin{pmatrix}g(x_{1},\ldots ,x_{n})x_{1}\\\vdots \\g(x_{1},\ldots ,x_{n})x_{n}\end{pmatrix}}}$

beschrieben. Die partiellen Ableitungen der ${\displaystyle {}j}$-ten Komponentenfunktion nach ${\displaystyle {}x_{i}}$ sind bei ${\displaystyle {}j=i}$ gleich

${\displaystyle {}{\frac {\partial }{\partial x_{i}}}{\left(g(x_{1},\ldots ,x_{n})x_{i}\right)}={\frac {\partial g}{\partial x_{i}}}(x_{1},\ldots ,x_{n})\cdot x_{i}+g(x_{1},\ldots ,x_{n})={\frac {\partial g}{\partial x_{i}}}(P)\cdot x_{i}+g(P)\,}$

und bei ${\displaystyle {}j\neq i}$ gleich

${\displaystyle {}{\frac {\partial }{\partial x_{i}}}{\left(g(x_{1},\ldots ,x_{n})x_{j}\right)}={\frac {\partial g}{\partial x_{i}}}(x_{1},\ldots ,x_{n})\cdot x_{j}={\frac {\partial g}{\partial x_{i}}}(P)\cdot x_{j}\,.}$
2. Das totale Differential ist
${\displaystyle {\begin{pmatrix}{\frac {\partial g}{\partial x_{1}}}(P)\cdot x_{1}+g(P)&{\frac {\partial g}{\partial x_{2}}}(P)\cdot x_{1}&\ldots &{\frac {\partial g}{\partial x_{n}}}(P)\cdot x_{1}\\{\frac {\partial g}{\partial x_{1}}}(P)\cdot x_{2}&{\frac {\partial g}{\partial x_{2}}}(P)\cdot x_{2}+g(P)&\ldots &{\frac {\partial g}{\partial x_{n}}}(P)\cdot x_{2}\\\vdots &\vdots &\ddots &\vdots \\{\frac {\partial g}{\partial x_{1}}}(P)\cdot x_{n}&{\frac {\partial g}{\partial x_{2}}}(P)\cdot x_{n}&\ldots &{\frac {\partial g}{\partial x_{n}}}(P)\cdot x_{n}+g(P)\end{pmatrix}}.}$