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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.

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Remember that $\dfrac{dy}{dx}$ is defined as some kind of limit of $\dfrac{\Delta y}{\Delta x}$ as $\Delta y$ and $\Delta x$ approach zero. So have a think about why this identity is true, and this may lead you to an answer to your own question. –  user22805 Aug 21 '12 at 13:01
It's true for partial derivatives, if you keep the same variables constant. Remember a partial derivatives needs to specify what you keep constant. –  Gerenuk Aug 21 '12 at 13:17
I suggest you abandon this medieval language and start thinking in terms of functions and their inverses. Variables are not functions, and derivatives are not fractions. –  Siminore Aug 21 '12 at 13:37
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2 Answers 2

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.

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thank you, nbubis –  emory j Aug 21 '12 at 14:09
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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 ${dx\over dy}$.

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thank you, christian –  emory j Aug 21 '12 at 14:09
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