How would one apply this to, say, the

__slope__ of a parameterized curve $\displaystyle \alpha=\arctan\left(\frac{\dot{y}}{\dot{x}}\right)$ at the point $\displaystyle (x,y)$?

$\displaystyle \frac{d}{dt} \Big{(}\alpha\Big{)} = \frac{d}{dt} \left(\arctan(\frac{ \dot{y} }{ \dot{x} })\right)=\frac{1}{\dot{x}^2+\dot{y}^2} \cdot \frac{d}{dt}\Big(\frac{\dot{y}}{\dot{x}}\Big{)} = \hdots$

Now, what is $\displaystyle \frac{d}{dt}\Big(\frac{\dot{y}}{\dot{x}}\Big{)}$?

The final answer should be $\displaystyle d\alpha=\frac{\dot{x}\ddot{y}-\ddot{x}\dot{y}}{\dot{x}^2+\dot{y}^2}dt$, so the $\displaystyle \frac{d}{dt}\Big(\frac{\dot{y}}{\dot{x}}\Big{)}$ apparently equals $\displaystyle \dot{x}\ddot{y}-\ddot{x}\dot{y}$. How do you get this?