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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 ?

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 ?

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.

Marine Galantin
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  • Related: is $\frac{dy}{dx}$ not a ratio? – JMoravitz Feb 13 '20 at 18:55
  • Also related: Inverse function theorem. – JMoravitz Feb 13 '20 at 18:56
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    There is some function $b \mapsto a(b)$. If it is differentiable at $b^$ and the derivative $a'(b^) \neq 0$ then the inverse function theorem says there is a local inverse function, call it $y$ such that $y(a(b)) = b$ for $b$ near $b^*$. The inverse function tells us that $y'(a(b)) = {1 \over a'(b)}$. This is where the formula is coming from. – copper.hat Feb 13 '20 at 19:00
  • @copper.hat thank you, that s a good point. I haven't thought about that theorem. – Marine Galantin Feb 13 '20 at 19:08
  • @WillJagy oh ok I see, sorry for that. I didn't mean to annoy you. It is just that I am used to prove a general statement, and if I struggle to do so, I ll try some example. I guess, trying to find a general rule from a particular case is a more advanced way of studying, but certainly more right and deeper. I wasn't thinking about my question that way (lazy Marine) but you're right – Marine Galantin Feb 13 '20 at 19:24

3 Answers3

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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.

2'5 9'2
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  • please see the edit – Marine Galantin Feb 13 '20 at 19:07
  • Even more generally, suppose $f(a,b)=0$. Then, under suitable smoothness conditions, we have $$\frac{\partial f}{\partial a},da+\frac{\partial f}{\partial b},db=0$$It is clear that $\frac{da}{db}$ is the reciprocal of $\frac{db}{da}$. – Mark Viola Feb 13 '20 at 19:28
  • Hi Alex. It seems that the OP is asking an important question, namely "Under what conditions does the equality hold?" In fact, the title asks "When is it true that …?" And the OP states the belief that $a$ and $b$ must at least be differentiable with respect to one another. So, I am not sure how an example helps answer the question in the title. – Mark Viola Feb 13 '20 at 19:41
  • @MarkViola The original question had this in it: "How can this be true ???". I saw that and interpreted it as OP expressing skepticism about the relation, and I was responding to that. I acknowledge now that is not really the question. – 2'5 9'2 Feb 13 '20 at 20:44
  • I changed my answer to something totally different. – 2'5 9'2 Feb 13 '20 at 22:35
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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 $y:=b$.

Christian Blatter
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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.

J.G.
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