Chain Rule: Is the notation $\frac{dy}{du} \cdot \frac{du}{dx} = \frac{dy}{dx}$ accurate?

Question

My question is if it is okay / mathematically rigorous to write the Chain Rule like that (the Leibniz way). I thought that $dx$, etc. do not follow the rules of algebra and cannot be treated as such. For example, I write $\int 1\, dx$, rather than $\int 1 \,dx$, and I write $\int \frac{dx}{a}$ instead of $\int \frac{1}{a}\, dx$.

So, is it correct to say $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ and, in essence, have the $\frac{du}{du}$ cancel to $1$?

(If my notion of $dx$ is not correct, I would also like an explanation of what really that is)

edit: Is it the same deal with $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$?

Answer 1

The notation $\frac{dy}{dx} = \frac{dy}{du} * \frac{du}{dx}$ is valid. However, you cannot prove the chain rule just by "cancelling" the two $du$'s; it doesn't work that way.

Answer 2

In terms of differentials (in the single-variable case), $\frac{dy}{dx}$ is the unique scalar with the property that $\frac{dy}{dx}dx = dy$.

$\frac{dy}{du} \frac{du}{dx}$ therefore has the property that

$$\frac{dy}{du} \frac{du}{dx} dx = \frac{dy}{du} du = dy $$

therefore $\frac{dy}{du}\frac{du}{dx} = \frac{dy}{dx}$.

You can't really justify the result by rearranging the expression like you would with fractions (e.g. by combining them into a single 'fraction'), which is why people mean when they say things like "you can't just cancel them". However, you can still prove (again this only makes sense in the single-variable case) that rearrangements are equal: e.g.

$$ \frac{dw}{dx} \frac{dy}{dz} = \frac{dw}{dz} \frac{dy}{dx} $$

(note that you could use this identity to prove your identity, because $\frac{du}{du} = 1$)

Answer 3

There is a $C$ which looks like $A/B$ (suppose) in some german notation. And now in some rule you denote some value by $C$ and others by $A$ and $B$ and $C = A/B$ is the rule. You won't say you broke the notation $C$ into $A/B$ because it was looking so.

You will deal more such cases in integration.

Answer 4

The relation $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$

requires that $u$ has a nonzero relationship with $x$ and $y$. This doesn't have to be explicitly stated when the chain rule is written as $$D_x f(g(x)) = f'g(x) \cdot g'(x)$$

Example:

Consider using $g = \text{gravitational force}$ and $r = \text{distance}$. You might know:

$$g = G\frac{m_1m_2}{r^2}$$

What if $u$ is your temperature? Perhaps your temperature does not change based on your location. $$\frac{du}{dr} = 0$$ $$\frac{dg}{du} = 0$$

If you applied the chain rule without realizing that $u$ is an independent variable, you'd get:

$$\begin{align}\frac{dg}{dr} &= \frac{dg}{du} \cdot \frac{du}{dr}\\ &= 0 \cdot 0 \end{align}$$

...and I think we can agree that gravitational force is not universally constant. So when using the differential form of the chain rule, make sure you are not using an independent variable.