Jacobimatrix/sin x durch x^2+y^4, x^2y durch x^2+y^2, ln (x^2+y^2)/Aufgabe/Lösung

Die partiellen Ableitungen sind

${}{\frac {\partial f_{1}}{\partial x}}={\frac {(x^{2}+y^{4})\cos x-2x\sin x}{(x^{2}+y^{4})^{2}}}\,,$

${}{\frac {\partial f_{1}}{\partial y}}={\frac {-4y^{3}\sin x}{(x^{2}+y^{4})^{2}}}\,,$

${}{\frac {\partial f_{2}}{\partial x}}={\frac {2xy(x^{2}+y^{2})-x^{2}y(2x)}{(x^{2}+y^{2})^{2}}}={\frac {2xy^{3}}{(x^{2}+y^{2})^{2}}}\,,$

${}{\frac {\partial f_{2}}{\partial y}}={\frac {x^{2}(x^{2}+y^{2})-x^{2}y(2y)}{(x^{2}+y^{2})^{2}}}={\frac {x^{4}-x^{2}y^{2}}{(x^{2}+y^{2})^{2}}}\,,$

${}{\frac {\partial f_{3}}{\partial x}}={\frac {2x}{x^{2}+y^{2}}}\,$

und

${}{\frac {\partial f_{3}}{\partial y}}={\frac {2y}{x^{2}+y^{2}}}\,.$

Somit ist die Jacobi-Matrix in einem Punkt ${}(x,y)$ gleich

${\begin{pmatrix}{\frac {(x^{2}+y^{4})\cos x-2x\sin x}{(x^{2}+y^{4})^{2}}}&{\frac {-4y^{3}\sin x}{(x^{2}+y^{4})^{2}}}\\{\frac {2xy^{3}}{(x^{2}+y^{2})^{2}}}&{\frac {x^{4}-x^{2}y^{2}}{(x^{2}+y^{2})^{2}}}\\{\frac {2x}{x^{2}+y^{2}}}&{\frac {2y}{x^{2}+y^{2}}}\end{pmatrix}}.$

Zur gelösten Aufgabe