noch eine Variante:
$$ y(t)=-4t^4+8t^2-1$$
$$ x(t)=t^2 \quad $$
$$ \frac {d \, y(t)}{d\, t}=-16t^3+16t$$
$$ \frac {d \,x(t)}{d\, t}=2t \quad $$
$$ \frac {d \,y(t)}{d\, x(t)}= \frac { \frac {d \,y(t)}{d\, t}}{ \frac {d \,x(t)}{d\, t}}$$
$$ \frac {d \,y(t)}{d\, x(t)}= \frac { -16t^3+16t}{2t}$$
$$ \frac {d \,y}{d\, x}= { -8t^2+8}$$
$$ \frac {d \,y}{d\, x}= { -8 x +8}$$