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

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Die relevanten Ableitungen sind

${\displaystyle {}{\frac {\partial f}{\partial x}}=e^{x}yz^{2}-y\,,}$
${\displaystyle {}{\frac {\partial f}{\partial y}}=e^{x}z^{2}-x\,,}$
${\displaystyle {}{\frac {\partial f}{\partial z}}=2e^{x}yz\,,}$
${\displaystyle {}{\frac {\partial ^{2}f}{\partial ^{2}x}}=e^{x}yz^{2}\,,}$
${\displaystyle {}{\frac {\partial ^{2}f}{\partial ^{2}y}}=0\,,}$
${\displaystyle {}{\frac {\partial ^{2}f}{\partial ^{2}z}}=2e^{x}y\,,}$
${\displaystyle {}{\frac {\partial }{\partial x}}{\frac {\partial f}{\partial y}}=e^{x}z^{2}-1\,,}$
${\displaystyle {}{\frac {\partial }{\partial x}}{\frac {\partial f}{\partial z}}=2e^{x}yz\,,}$
${\displaystyle {}{\frac {\partial }{\partial y}}{\frac {\partial f}{\partial z}}=2e^{x}z\,.}$

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

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

Daher ist das Taylor-Polynom der Ordnung zwei gleich

${\displaystyle (e-1)y+(e-1)(x-1)y-2ey(z+1).}$