9
$\begingroup$

We all know that since $\textrm{d}y$ and $\textrm{d}x$ are not numbers, we can't treat $\frac{dy}{dx}$ as a ratio. However, in various operations, this treatment seems to work. I've picked a few examples:

  • Chain Rule: $$\frac{du}{dx} = \frac{du}{dy} \frac{dy}{dx}$$
  • Variable separation: $$ \frac{dy}{dx} = f(x)h(y)$$ $$\frac{dy}{h(y)} = f(x)dx$$
  • Integral of $\frac{dy}{dx}$: $$ \int \frac{dy}{dx}dx = \int dy = y$$ And there is also some examples that this treatment does not work: $$\frac{dy}{dx} = - \frac{\partial F / \partial y}{\partial F / \partial x}$$ Where the minus sign shows that this is a counterexample. Now, if this isn't correct, why does it works for most cases? And, if it works for most cases, why there are a few cases that it won't work? Thanks.

References

1) Answer from asmeurer in Is $\frac{dy}{dx}$ not a ratio?

$\endgroup$
9
$\begingroup$

When Leibniz designed the notation he was thinking of $\frac {dy}{dx}$ as a ratio of infinitesimals. But, the analysts of the 18th century were unable to develop this into a rigorous theory, and infinitesimals were replaced by limits.

And $\frac {dy}{dx}$ no longer a ratio.

However, it does work like a ratio for just about all of elementary calculus.

When you get multivariate calculus and partial differentiation, the thought of the differential operator as a ratio really starts to break down.

the total derivative: $\frac {d}{dt} f(u,v) = \frac {\partial f}{\partial u} \frac {du}{dt} + \frac {\partial f}{\partial v} \frac {dv}{dt}$

And as the math becomes increasingly advanced the differential operator behaves less and less like a ratio.

$\endgroup$
7
$\begingroup$

Note, that in the counterexample you gave, $F$ is a function of $y$ and $x$. Thus, $\frac{\partial F}{\partial y}$ is a shorthand for $\lim\limits_{h \to 0} \frac{F(x,y+h) - F(x,y)}{h}$ which is very much different than $\lim\limits_{h \to 0} \frac{F(x+h,y) - F(x,y)}{h}$.

For this reason you can't cancel $\partial F$ against each other, thus:

$$\frac{\partial F/ \partial y }{\partial F/ \partial x} \neq \frac{\partial x}{\partial y}$$

$\endgroup$
4
$\begingroup$

Here's a nice example: Consider the equation $xyz=1.$ Then

$$\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}\frac{\partial z}{\partial x} = -1,$$

even though cancellation of the differentials suggests otherwise.

$\endgroup$
2
$\begingroup$

There is a simple answer: $y'(x)$ is in fact approximately equal to $\Delta y/\Delta x$, where $\Delta x$ is an extremely tiny (but finite) change in the value of $x$ and $\Delta y$ is the corresponding change in the value of $y$. Most informal derivations involving the "infinitesimals" $dx$ and $dy$ can easily be rephrased in terms of $\Delta x$ and $\Delta y$, with the equals signs replaced by $\approx$ signs. This only shows that the equation you derive is approximately true, but it seems plausible that you can make the approximation as good as you like by using a sufficiently small value of $\Delta x$. If so, then it follows that the equation you derived must hold with exact equality.

(The "it seems plausible" step is where we are likely to get into trouble if we aren't careful.)

$\endgroup$
2
$\begingroup$

$\mathrm{d}x$ and $\mathrm{d}y$ aren't numbers, but they are differentials. And differentials can be multiplied by scalars.

And if $\mathrm{d}x$ is nonsingular ($x$ being a so-called "independent variable" guarantees this), then if $z \, \mathrm{d}x = \mathrm{d} y $ has any solutions for $z$, it has a unique solution.

And thus, $z$ really is the ratio of $\mathrm{d}y$ to $\mathrm{d}x$, and it makes sense to say $z = \frac{\mathrm{d}y}{\mathrm{d}x}$.


The usual notation for partial differentiation is... bad. Partial derivatives aren't like ratios at all. Don't treat them like ratios.

More accurately, partial derivatives only become ratios after making a bunch of modifications to the formula being differentiated... and the notation doesn't even tell you what precisely those modifications are supposed to be; you have to infer from context.

For example, if the differential of some bivariate formula is

$$ \mathrm{d}F(x,y) = g(x,y) \mathrm{d} x + h(x,y) \mathrm{d} y $$

then what the notation $\partial F(x,y) / \partial x$ means is:

  • First infer that one means to single out $x$ and $y$ as "independent" variables.
  • Restrict all the "independent" variables other than $x$ to be constant — in particular, you have $\mathrm{d}y = 0$.
  • Simplify the equation to $\mathrm{d}F(x,y) = g(x,y) \mathrm{d} x$.
  • Now, there really is a ratio between the two differentials. Define $\partial F(x,y)/\partial x = g(x,y)$ to be that ratio.

Actually, rather than the last two steps, it's more accurate to say that $\partial / \partial x$ means to get to additionally apply to $\mathrm{d}F(x,y)$ the transformation $\mathrm{d}x \mapsto 1$ alongside $\mathrm{d}y \mapsto 0$, but I wanted to give a description that at most resembled a ratio.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.