fina typsätningar A.Q. och Susanne

Era labbar är godkända!

mvh 😊

Funktionen \[\frac {\left (1 - x^2 \right) \left (y^2 - 4 \right) - x^2 - y^2 +
5}{\left (x^2 + y^2 + 1 \right)^2}\]

M.a.p:x
\[\frac {-2 x\left (y^2 - 4 \right) -
2 x}{\left (x^2 + y^2 +
1 \right)^2} - \frac {4 x\left (\left (1 -
x^2 \right)\left (y^2 - 4 \right) - x^2 - y^2 +
5 \right)}{\left (x^2 + y^2 + 1 \right)^3}\]
M.a.p:y
\[\frac {2\left (1 - x^2 \right) y -
2 y}{\left (x^2 + y^2 +
1 \right)^2} - \frac {4 y\left (\left (1 -
x^2 \right)\left (y^2 - 4 \right) - x^2 - y^2 +
5 \right)}{\left (x^2 + y^2 + 1 \right)^3}\]

\[
f(x,y)=\frac{\left(1-x^2\right) \left(y^2-4\right)-x^2-y^2+5}{\left(x^2+y^2+1\right)^2}
\]

\[
\frac{df}{dx}=\frac{-2 x \left(y^2-4\right)-2 x}{\left(x^2+y^2+1\right)^2}-\frac{4 x \left(\left(1-x^2\right) \left(y^2-4\right)-x^2-y^2+5\right)}{\left(x^2+y^2+1\right)^3}
\]

\[
\frac{df}{dy}=\frac{2 \left(1-x^2\right) y-2 y}{\left(x^2+y^2+1\right)^2}-\frac{4 y \left(\left(1-x^2\right) \left(y^2-4\right)-x^2-y^2+5\right)}{\left(x^2+y^2+1\right)^3}
\]