Einheitskreis/Obere Halbebene/Rational isomorph/Reelle partielle Ableitungen/Aufgabe/Lösung

Die partiellen Ableitungen von

${}\psi {\begin{pmatrix}a\\b\end{pmatrix}}={\frac {1}{-2a+1+a^{2}+b^{2}}}{\begin{pmatrix}-2b\\1-a^{2}-b^{2}\end{pmatrix}}\,$

sind

${}{\begin{aligned}{\frac {\partial \psi }{\partial a}}&={\frac {1}{{\left(-2a+1+a^{2}+b^{2}\right)}^{2}}}{\begin{pmatrix}2b{\left(2a-2\right)}\\-2a{\left(-2a+1+a^{2}+b^{2}\right)}-{\left(1-a^{2}-b^{2}\right)}{\left(-2+2a\right)}\end{pmatrix}}\\&={\frac {1}{{\left(-2a+1+a^{2}+b^{2}\right)}^{2}}}{\begin{pmatrix}2b{\left(2a-2\right)}\\4a^{2}-4a+2{\left(1-a^{2}-b^{2}\right)}\end{pmatrix}}\\&={\frac {1}{{\left(-2a+1+a^{2}+b^{2}\right)}^{2}}}{\begin{pmatrix}4ab-4b\\2a^{2}-4a+2-2b^{2}\end{pmatrix}}\end{aligned}}$

und

${}{\begin{aligned}{\frac {\partial \psi }{\partial b}}&={\frac {1}{{\left(-2a+1+a^{2}+b^{2}\right)}^{2}}}{\begin{pmatrix}-2{\left(-2a+1+a^{2}+b^{2}\right)}+4b^{2}\\-2b{\left(-2a+1+a^{2}+b^{2}\right)}-2b{\left(1-a^{2}-b^{2}\right)}\end{pmatrix}}\\&={\frac {1}{{\left(-2a+1+a^{2}+b^{2}\right)}^{2}}}{\begin{pmatrix}4a-2-2a^{2}+2b^{2}\\4ab-4b\end{pmatrix}}.\end{aligned}}$

Zur gelösten Aufgabe