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I want to prove $y'_x=\frac{1}{x'_y}$. $y'_x$ means the derivative of $y$ with respect to $x$ and $x'_y$ means the derivative of $x$ with respect to $y$.

So I know $y'_x=\frac{dy}{dx}$ and $x'_y=\frac{dx}{dy}$. My main question is, can we consider $\frac{dy}{dx}$ and $\large\frac{dx}{dy}$ as typical fractions like $\frac{a}{b}$ $(b\neq0)$? Because then it would be easy to prove that.

If I'm not mistaken, my professor said that these are not fractions. She said in the chain rule, where we have two functions $y=f(u)$ and $u=g(x)$, we write $\frac{dy}{dx}=\frac{d}{dx}\bigl(f(u)\bigr)=\frac{dy}{du}\cdot\frac{du}{dx}$ so now you want to say we cancel both $du$s: $\require{cancel}\frac{dy}{\cancel{du}}\cdot\frac{\cancel{du}}{dx}$ and she said it is wrong. Because these aren't fractions.

But when I think about it, I feel that it makes sense to consider these fractions. Why is that? Because something like $dx$ or $dy$ is basically the difference between two values of $x$ or $y$ (in the derivative, it's an infinitesimal change), so it's like dividing them.

Xander Henderson
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  • You can at least make an intuitive argument. If $\Delta x$ is a tiny (but finite) number and $\Delta y = y(x + \Delta x) - y(x)$, then $\frac{dy}{dx} \approx \frac{\Delta y}{\Delta x} = \frac{1}{\frac{\Delta x}{\Delta y}}$. You might hope "in the limit" to obtain an exact equality. This type of "infinitesimal intuition" is very useful and can make calculus seem easy. Real analysis textbooks show how to develop calculus with airtight logic, and that is nice, but we should not throw infinitesimal intuition out the window. – littleO Dec 01 '24 at 18:30

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