Taylorpolynom/Zweite Ordnung/e^x yz^2-xy/(1,0,-1)/Aufgabe/Lösung

Die relevanten Ableitungen sind

${}{\frac {\partial f}{\partial x}}=e^{x}yz^{2}-y\,,$

${}{\frac {\partial f}{\partial y}}=e^{x}z^{2}-x\,,$

${}{\frac {\partial f}{\partial z}}=2e^{x}yz\,,$

${}{\frac {\partial ^{2}f}{\partial ^{2}x}}=e^{x}yz^{2}\,,$

${}{\frac {\partial ^{2}f}{\partial ^{2}y}}=0\,,$

${}{\frac {\partial ^{2}f}{\partial ^{2}z}}=2e^{x}y\,,$

${}{\frac {\partial }{\partial x}}{\frac {\partial f}{\partial y}}=e^{x}z^{2}-1\,,$

${}{\frac {\partial }{\partial x}}{\frac {\partial f}{\partial z}}=2e^{x}yz\,,$

${}{\frac {\partial }{\partial y}}{\frac {\partial f}{\partial z}}=2e^{x}z\,.$

Somit sind die Werte der relevanten Ableitungen im Punkt ${}(1,0,-1)$ gleich

${}f(1,0,-1)=0\,,$

${}{\frac {\partial f}{\partial x}}(1,0,-1)=0\,,$

${}{\frac {\partial f}{\partial y}}(1,0,-1)=e-1\,,$

${}{\frac {\partial f}{\partial z}}(1,0,-1)=0\,,$

${}{\frac {\partial ^{2}f}{\partial ^{2}x}}(1,0,-1)=0\,,$

${}{\frac {\partial ^{2}f}{\partial ^{2}y}}(1,0,-1)=0\,,$

${}{\frac {\partial ^{2}f}{\partial ^{2}z}}(1,0,-1)=0\,,$

${}{\frac {\partial }{\partial x}}{\frac {\partial f}{\partial y}}(1,0,-1)=e-1\,,$

${}{\frac {\partial }{\partial x}}{\frac {\partial f}{\partial z}}(1,0,-1)=0\,,$

${}{\frac {\partial }{\partial y}}{\frac {\partial f}{\partial z}}(1,0,-1)=-2e\,.$

Daher ist das Taylor-Polynom der Ordnung zwei gleich

$(e-1)y+(e-1)(x-1)y-2ey(z+1).$

Zur gelösten Aufgabe