Identität/Skalare Funktion/Ableitung/Aufgabe/Lösung

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

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

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

und bei ${}j\neq i$ gleich

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