When is it true that $\frac{ \partial a } { \partial b} = (\frac{ \partial b } { \partial a} )^{-1} $

Question

My statistics' lecturer wrote that weird sentence :

$$\frac{ \partial a } { \partial b} = (\frac{ \partial b } { \partial a} )^{-1} $$

How can this be true ??? Does anyone know under what condition, an analyst would agree with tha statement ?

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First I would say that $a$ and $b$ have to be differentiable with respect to each other. But that sentence seems so weird. Does that mean that one of them needs the existence of the inverse ?

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EDIT:

I see most answers for now are about examples. I am not saying that it doesn't make sense. I agree that most of non-mathematicians are seeing it as a ratio. I am asking under what conditions for $a$, $b$, can one write that.

Answer 1

When there are two variable quantities $a$ and $b$ that are tied to each other in a way that the value of any one of them determines the value of the other one, then we have (meaning: there are present) two functions $$a\mapsto b=\phi(a),\qquad b\mapsto a=\psi(b)\ ,$$ and one can say that $\phi$ and $\psi$ are inverses of each other in certain $a$- and $b$-domains. What your professor denotes as ${\partial b\over\partial a}$ is nothing else than $\phi'(a)$, and ${\partial a\over\partial b}$ is $\psi'(b)$.

Now it is proven in calculus 101 that the derivatives of inverse functions $\phi=\psi^{-1}$ in admissible points $(a,b)$ are connected by $$\phi'(a)\cdot\psi'(b)=1\ .$$ For a proof under the assumption that the derivatives exist, apply the chain rule to $$\phi\bigl(\psi(y)\bigr)\equiv y$$ at the point $y:=b$.

Answer 2

This is an informal explanation.

A derivative ($\frac{da}{db}$) is the limit of a ratio of very small things. Traditionally defined as the limit where the denominator approaches $0$. But keep in mind that for the limit to exist, as one of these quantities approaches $0$, the other one must also approach $0$. So it's effectively a limit as both $\Delta a$ and $\Delta b$ approach $0$.

With this in mind, then it is no surprise that the limit of the (multiplicative) inverse equals the (multiplicative) inverse of the limit. That is, that $$\lim\frac{\Delta a}{\Delta b}=\left(\lim\frac{\Delta b}{\Delta a}\right)^{-1}$$ In the above, the two "$\lim$" are traditionally limits with respect two different things. But my point above is that form a certain perspective they are the same.

Answer 3

This kind of partial derivative inversion is valid when the held-constant conditions that define these derivatives are the same. If we compare Cartesian coordinates with polar ones in $2$ dimensions, do you argue $\frac{\partial r}{\partial x}=\partial_x(r\sec\theta)=\sec\theta=\frac{r}{x}$, or $\frac{\partial r}{\partial x}=\partial_x\sqrt{x^2+y^2}=\frac{x}{\sqrt{x^2+y^2}}=\frac{x}{r}$? The first option holds $\theta$ constant, the second $y$ constant. Something similar happens when you work with Lorentz transformations. In fact, both example give results of the form $\left(\frac{\partial a}{\partial b}\right)_c=\left(\frac{\partial a}{\partial b}\right)_d^{-1}$, where the subscripted variable is held fixed.