Working with infinitesimals of the form d(f(x)), for example d(ax), and relating them to dx (integration, delta function) I am trying to get a better understanding on how we can manipulate the infinitesimal dx in an integral $$\int f(x) dx$$
I have come across the following
$$ d(\cos (x)) = -\sin(x) dx$$
Therefore
$$\int^{x=2\pi}_{x=0} dx \sin(x) \cos(x) = - \int^{x = 2\pi}_{x=0} d(\cos(x)) \cos(x) = - \dfrac{1}{2} [ \cos^{2}(x)]^{x=2\pi}_{x=0} = -\dfrac{1}{2}[1-1] = 0$$
This looks to me like the chain rule can be applied to infinitesimals in analogy to differentiation.
However, today I'm trying to solve the following problem : prove
$$\delta(ax) = \dfrac{\delta(x)}{|a|}$$
Following the hint I looked at  $$\int d(ax)\delta(ax) = 1 = \int d(ax)\delta(-ax)$$
Since $$\int d(ax)\delta(ax) = 1 \quad \text{and} \quad \delta(x) = \delta(-x)$$
From this it would seem
$$d(ax) = |a|dx$$
giving $$\int d(ax)\delta(ax) = |a|\int dx \delta(ax) = |a|\int dx \delta(-ax) = \int dx \delta(x) = 1$$
as expected.
I would have naively assumed $d(ax) = a \space dx$
In summary, I have no idea how to treat d(f(x)), and I'm not sure where to look for information. Could someone help me gain a better understanding ? Unfortunately I have only taken a few undergraduate maths courses so far, so I couldn't understand anything too complex.
 A: The answer of md2perpe is the good way to prove what you want to prove. Another way to solve your problem, is to remark that defining the Heaviside function $H = \mathbb{1}_{\mathbb{R}_+}$, one has $H' = \delta_0$ and $H(ax) = \mathrm{sign}(a)\,H(x)$. Therefore
$$
\begin{align*}
\delta_0(a\,x) &= H'(a\,x) = \frac{1}{a} \frac{\mathrm d}{\mathrm d x} (H(a\,x))
\\
&= \frac{1}{a} \frac{\mathrm d}{\mathrm d x} (\mathrm{sign}(a)\,H(x)) = \frac{1}{|a|} H'(x) 
\\
&= \frac{1}{|a|} \delta_0(x)
\end{align*}
$$

I will here add some comment about the notation $\mathrm d(f(x))$. One of the problems with this notation is that $\mathrm d x$ denotes the Lebesgue measure, while $\delta$ (which I prefer to write $\delta_0$) is not a Lebesgue measurable function but also a measure. So one should not use the expression
$$
∫ \delta_0(x) \,\mathrm{d} x
$$
but either $∫ f(x) \,\mathrm{d} x$
if $f$ is a Lebesgue measurable function, and $∫ f\,\delta_0 = f(0)$ if $f$ is a $\delta_0$ measurable function (e.g. a function continuous in $0$). In some sense, a measure is only defined on sets and not on points, so if we identify $\mathrm d x$ with the indication of a local volume, then we should rather write
$$
∫ f(x) \,\delta_0(\mathrm{d}x)
$$
An other good formalism is the one of the Stieltjes integral (see e.g. https://en.wikipedia.org/wiki/Lebesgue%E2%80%93Stieltjes_integration). In this formalism, if $g$ is a function of bounded variations, then one can define
$$
∫ f\,\mathrm{d}g = \int f(x)\,\mathrm{d}g(x)
$$
and actually, since $g$ is of bounded variations if and only if its derivative in the sense of distributions $g'$ is a measure. So, as a distribution, we have
$$
\langle g',f\rangle = ∫ f(x) \,\mathrm{d}g(x)
$$
(or if you do not know distributions, let say that if $g'$ is integrable then we have $\int f\,g' = ∫ f \,\mathrm{d}g$). So, to have coherent notations, one should write $∫ f\,\mathrm d g$ to indicate that one integrate with respect to the measure $g'$, and not $g$. For example, for the Dirac delta, this gives
$$
∫ f(x)\,\mathrm{d}H(x) = ∫ f(x)\,\delta_0(\mathrm{d}x) = \langle \delta_0,f\rangle = f(0)
$$
Here the first integral is well defined as a Lebesgue-Stieltjes integral, the second as an integral with respect to a measure and the third as a distribution.
A: Let $\varphi$ be a test function.
If $a>0$ then
$$
\int_{-\infty}^{\infty} \delta(ax) \, \varphi(x) \, dx 
= \{ y=ax \} 
= \int_{-\infty}^{\infty} \delta(y) \, \varphi(y/a) \, \frac{1}{a} dy \\
= \frac{1}{a} \varphi(0)
= \int_{-\infty}^{\infty} \frac{1}{a} \delta(x) \, \varphi(x) \, dx .
$$
If $a<0$ then
$$
\int_{-\infty}^{\infty} \delta(ax) \, \varphi(x) \, dx 
= \{ y=ax \} 
= \int_{\infty}^{-\infty} \delta(y) \, \varphi(y/a) \, \frac{1}{a} dy
= - \int_{-\infty}^{\infty} \delta(y) \, \varphi(y/a) \, \frac{1}{a} dy \\
= -\frac{1}{a} \varphi(0)
= \int_{-\infty}^{\infty} \frac{1}{-a} \delta(x) \, \varphi(x) \, dx .
$$
Thus, for any $a\neq 0,$
$$
\int_{-\infty}^{\infty} \delta(ax) \, \varphi(x) \, dx 
= \int_{-\infty}^{\infty} \frac{1}{|a|} \delta(x) \, \varphi(x) \, dx .
$$
Since this is valid for all test functions $\varphi$ we have
$$
\delta(ax) = \frac{1}{|a|} \delta(x).
$$
