What is this type of differentiation called?

Question

I understand that the result is found using chain rule, but I'm not sure why we are allowed to do this (differentiate with respect to, apparently, nothing?)

Example:

$$x = \frac{c}{2}(1 - \cos\theta)$$

becomes:

$$dx = \frac{c}{2}\sin\theta \cdot d \theta$$

Answer 1

You are not differentiating with respect to "nothing". Originally, you have

$$ x = \frac{c}{2}\left(1-\cos(\theta)\right).$$

If you differentiate $x$ with respect to $\theta$, it follows that:

$$ \frac{dx}{d\theta} = \frac{c}{2}\sin(\theta).$$

But this is equivalent to what you posted, i.e.,

$$ \frac{dx}{d\theta} = \frac{c}{2}\sin(\theta) \Leftrightarrow dx = \frac{c}{2}\sin(\theta)d\theta.$$

Answer 2

For some uses you write $\frac{dx}{d\theta}=f(\theta)\text{ as } dx=f(\theta) d\theta$ mostly if you want to integrate

Answer 3

My paper, "Total and Partial Differentials as Algebraically Manipulable Entities" (link to arXiv version) gives a justification for taking differentials in this way.

Essentially, let us suppose that all variables are dependent on some ultimately independent continuous variable, $q$. Therefore, if you say $x$ or $y$ you are really saying $x(q)$ or $y(q)$. The differential operator $d()$ evaluates the difference between the expression at $q = q + \epsilon$ and $q = q$, where $\epsilon$ is a fixed hyperreal infinitesimal. See Section 4.8 of the above paper.

As a definition, for any variable $u$, $du = d(u) = d(u(q)) = u(q + \epsilon) - u(q)$. Therefore, we can also say that $u(q + \epsilon) = du + u(q)$. We can use this to simplify generate the standard rules of calculus.

Example: $y = x^2$.

$$ y(q) = (x(q))^2 \\ d(y(q)) = d((x(q))^2) \\ d(y(q)) = (x(q + \epsilon))^2 - x(q)^2 \\ d(y(q)) = (d(x(q)) + x(q))^2 - x(q)^2 \\ d(y(q)) = d(x(q))^2 + 2x\,d(x(q)) + x(q)^2 - x(q)^2 \\ d(y(q)) = d(x(q))^2 + 2x\,d(x(q)) \\ d(y(q)) = 2x\,d(x(q)) \\ dy = 2x\,dx $$

Here, the next to last simplification utilizes the principle part of the hyperreal value (i.e., the most significant portion - since $dx^2$ is an order of infinity smaller that $dx$, it is ignored in the principle part), and the very last simplification simply reverts from the expanded notation to the simplified notation.

Answer 4

For all intents and purposes, you should think of $d$ as a derivative, but when it hits a variable (i.e. the $\theta$ in your example), it creates a new "variable" called $d\theta$. The purpose of such a thing is that these objects naturally come up in integration, namely in u-substitution. Let me give you a simple example. Consider

$$\int_0^2\cos(2\theta)d\theta;$$

in your first calculus course you learn to integrate such a function, one would need to perform the u-substitution $u=2\theta$ with $du=d(2\theta)=2d\theta$, so that $d\theta=\frac{du}{2}$. This last equality you should think about as a "correction term", as

$$\int_{x=0}^{x=2} \cos(2\theta)d\theta=\int_{u=0}^{u=4}\cos(u)\left(\frac{du}{2}\right).$$

Had I not written the $1/2$ in the RHS, the above equality of integrals would not be true, which is why I need such a correction term. Now what you've written down is simply a more complicated change of variables, but the same idea applies. Instead of a 1/2, your correction term is just a $c/2\sin\theta$, meaning for any function $f$, $$\int_{x=\frac{c}{2}(1-\cos a)}^{x=\frac{c}{2}(1-\cos b)} f(x)dx=\int_{\theta=a}^{\theta=b} f\left(\frac{c}{2}(1-\cos\theta)\right) \left(\frac{c}{2}\sin\theta d\theta\right).$$

Formally, this $d$ is called an "exterior derivative" and these "new variables"/"correction terms" are called 1-forms. It's essentially a whole bunch of machinery to formalize the process of u-sub.