"Reciprocal" of derivatives Let $x,y$ be 2 variables. 
When then is ${dx\over dy }= {1\over {dy\over dx}}$? I guess it is true for total derivatives, but am not entirely sure. 
What about if the derivatives are only partial derivatives? 
Many thanks.
 A: This is indeed true for total derivatives (assuming the derivatives are non-zero), The theorem stating this is called the "Inverse function theorem". For more than one variable, this involves Jacobians rather than derivatives, but the idea is the same.
A: The following setup should be sufficiently general: Tacitly underlying is an open set (a "phase space") $\Omega\subset{\mathbb R}^n$, and on $\Omega$ two scalar functions ("observables") $$X:\quad {\bf u}\mapsto X({\bf u})\ ,\qquad Y:\quad {\bf u}\mapsto Y({\bf u})$$
are defined. When the point ${\bf u}$ moves around in time according to some law $t\mapsto{\bf u}(t)\in\Omega$ then the "observables" $X$ and $Y$ become functions of time, too, and one is led to consider the functions
$$x(t):=X\bigl({\bf u}(t)\bigr)\ ,\qquad y(t):=Y\bigl({\bf u}(t)\bigr)\ .$$
By the chain rule these functions have derivatives
$$\dot x(t)=\nabla X\bigl({\bf u}(t)\bigr)\cdot{\bf u}'(t)\ ,$$
and similarly for $y(\cdot)$. At any given instant the quantity
$${dy\over dx}:={\dot y(t)\over \dot x(t)}={\nabla Y\bigl({\bf u}(t)\bigr)\cdot{\bf u}'(t) \over \nabla X\bigl({\bf u}(t)\bigr)\cdot{\bf u}'(t)}$$
is obviously the reciprocal of ${dx\over dy}$.
