Examples where the product of reciprocal derivatives isn't one?

Question

I'm trying to understand why it isn't a good idea to treat derivatives like fractions. Could someone give me an example of a function $y$ such that $$\frac{dy}{dx} \cdot \frac{dx}{dy} \not = 1$$

This inspired my question, so I would appreciate it if someone could address it:

Conside a function $PV = kT$ for some constant $k$. Then $\dfrac{\partial P}{\partial V}\dfrac{\partial T}{\partial P}\dfrac{\partial V}{\partial T} = -1$ instead of $1$ as you would have expected if it's valid to treat derivatives "intuitively" as fractions.

Answer 1

The "cancellation of fractions" is really a consequence of the inverse function theorem:

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Inverse Function Theorem Let $f:\mathbb{R}^n\to \mathbb{R}^m$ be differentiable. Suppose that the Jacobian matrix of $f$, $Df(x)$ is invertible for some $x$. Then $f$ is locally invertible, and its inverse $f^{-1}$ satisfies $$D(f^{-1})(f(x))=Df(x)^{-1}$$

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In the single-variable case, i.e. for $f:\mathbb{R}\to \mathbb{R}$ differentiable, then the Jacobian of $f$ is given by the $1\times 1$ matrix $$Df(x)=\begin{bmatrix}\frac{\mathrm{d}f}{\mathrm{d}x}\bigg|_x\end{bmatrix}$$ Remember that a linear map from $\mathbb{R}\to \mathbb{R}$ is invertible if and only if it is non-zero. So the inverse function theorem takes the following form:

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Inverse Function Theorem (Single variable calc) Let $f:\mathbb{R}\to \mathbb{R}$ be differentiable. Suppose that $\frac{\mathrm{d}f}{\mathrm{d}x}\big|_x\neq 0$ for some $x$. Then $f$ is locally invertible, and its inverse $f^{-1}$ satisfies $$\begin{bmatrix}\frac{\mathrm{d}f^{-1}}{\mathrm{d}x}\bigg|_{f(x)}\end{bmatrix}=\begin{bmatrix}\frac{\mathrm{d}f}{\mathrm{d}x}\bigg|_x\end{bmatrix}^{-1}$$

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But the inverse of a $1\times 1$ matrix is just $$\begin{bmatrix}a\end{bmatrix}^{-1}=\begin{bmatrix}\frac{1}{a}\end{bmatrix}$$ and so, letting $~y=f(x)$, $$\frac{\mathrm{d}x}{\mathrm{d}y}\bigg|_{y}\frac{\mathrm{d}y}{\mathrm{d}x}\bigg|_x=1.~~\square$$

Answer 2

An intuitive way to see why $$\frac{dx}{dy}\frac{dy}{dx}=1$$ would be if we treat the ratio as if it were actually a fraction. In particular, suppose that $x$ and $y$ are actually functions of some dummy variable - we could imagine, for instance, that they take on various values over time - but they happen to always satisfying some relation. Like, for instance, we could have $x$ be the volume of a balloon and $y$ be the pressure within. The two variables depend on time, but the quantity $\frac{dx}{dy}$ still makes sense - and can be interpreted equivalently either as the change in volume given a change in pressure OR literally as the ratio of those two derivatives where $dx$ is taken to be the derivative of $x$ with respect to the dummy variable time.

That this works is equivalent to the rule for composition of functions - that is the property that $(f\circ g)'(t)=f'(g(t))g'(t)$. In particular, suppose we had $x$ and $y$ related by $f$ so $x=f(y)$. Then, we take the derivative of both sides, where we write the derivative of $x$ as $dx$. Then we get: $$dx = f'(y) dy$$ or, in other words, where $x$ and $y$ are functions of $t$ $$x'(t)=f'(y(t))y'(t)$$ by the rule for composition; note that we aren't differentiating with respect to any particular variable, and yet if we considered both $x$ and $y$ to be functions of their own, the values $dx$ and $dy$ are in fact well-defined, being simply the derivatives. So, we interpret the quantity $$\frac{dx}{dy}=f'(y)$$ to read that we literally want the ratio of those two derivatives - and the statement above: "$\frac{dx}{dy}$ is the derivative of the function relating $x$ to $y$" is a consequence of the product rule. Thus, we can legally cancel in the fractions, so long as neither derivative is $0$.

The same technique, unfortunately doesn't work for partial fractions, since if we tried to write out $$x = f(y,z)$$ and differentiated to $$dx = f_y(y,z)dy + f_z(y,z)dz$$ we would see that dividing by $dy$ or $dz$ would yield an expression dependent on either $\frac{dy}{dz}$ or $\frac{dz}{dy}$; the problem is that the above interpretation of derivatives only works well in a single variable, since it implicitly assumes that every variable is a function of a single dummy variable.

Answer 3

Regarding the OP's additional comments concerning partial derivatives, the multivariable chain rule exhibits quite clearly the absurdity of treating partial derivatives blindly as fractions.

Consider a composition of function $$\mathbb{R}^2 \xrightarrow{f} \mathbb{R}^2 \xrightarrow{g} \mathbb{R}^2 $$ Inserting variables $(s,t)$ for the first $\mathbb{R}^2$, variables $(x,y)$ for the middle $\mathbb{R}^2$, and variables $(z,w)$ for the last $\mathbb{R}^2$, and writing coordinate functions for $f=(f_1,f_2)$ and for $g=(g_1,g_2)$, we can write this composition as $$x = f_1(s,t), \quad y = f_2(s,t) $$ $$z = g_1(x,y), \quad w = g_2(x,y) $$ Assuming differentiability of the functions $f$ and $g$, the multivariable chain rule gives us a total of four equations expressing the partials of each of $z,w$ with respect to each of $s,t$. Here is just one of those four equations: $$\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s} $$ And now, if you are in a free-wheeling mood, cancel and add to get $$\frac{\partial z}{\partial s} = 2 \frac{\partial z}{\partial s} $$