I was asked to solve such an (integral: $fig. 1$), where the derivative of the delta-function is made with respect to his argument (i guess something like: $fig. 2$). But, let's focus only on my attempt to evaluate that derivative: i did the explicit calculus making use of integration by parts, ending up with the (following: $fig. 3 \Rightarrow fig. 4$). Substituting the last result in the original integral, i (get: $fig. 5$). This is where i had the first doubt... Is the step ($fig. 5 \Rightarrow fig. 3$) legit, regardless the changing of differential terms ($dg \Rightarrow dx$)? Anyway, i am confused about the meaning of $\left[\frac{d}{dg} f(x)\right]$. In other situations i would say $\left[\frac{d}{dg} f(x) = \frac{f'(x)}{g'(x)}\right]$, that was obtained by treating each differential term $(df, dg, dx)$ like "independent-numbers" in leibniz's way; but in this specific situation, i think i should treat each term in an operatorial way, preserving the order of derivatives (see: $fig.6$).
REAL QUESTION: (to me) the last intuition is no more that an "intuition" based on the solutions i have for some exercises by my textbook; however this still remains a big doubt, since i wasn't able to do any real reasonement. It would be very appreciated if anyone could suggest how to demonstrate this.
$$\int_{a}^{b}\delta'[g(x)]f(x) \, dx \qquad (1)$$
$$\delta'[g(x)]=\frac{d}{dg}\delta[g(x)] \qquad (2)$$
$$ \require{cancel} \int_{a}^{b}\delta'[g(x)]\, f(x) \, dg = \cancelto{0} {\delta[g(x)]f(x)\vert_{a}^b} - \int_{a}^{b} \delta[g(x)]\, \frac{d}{dg} f(x) \, dg \qquad (3)$$
$$\delta'[g(x)]=-\delta[g(x)]\frac{d}{dg} \qquad (4)$$
$$\int_{a}^{b} \delta'[g(x)]f(x) \, dx = -\int_{a}^{b} \delta[g(x)]\, \frac{d}{dg} f(x) \, dx\qquad (5)$$
$$\left[\frac{d}{dg}\right]f(x) = \left[\frac{d}{dx} \cdot \frac{dx}{dg}\right]f(x) = \left[\frac{d}{dx} \cdot \left(\frac{dg}{dx}\right)^{-1}\right]f(x) = \frac{d}{dx}\left[\frac{f(x)}{g'(x)}\right] \qquad (6)$$
The main concern in your developments is the notation; indeed, as is, the notation $\delta'(g(x))$ can have three meanings :
the derivative of the Dirac delta with respect to $g$, which is itself reparametrized thanks to the variable $x$, i.e. $\delta'(g(x)) = \frac{\mathrm{d}\delta}{\mathrm{d}g}$ with $g = g(x)$;
the derivative of the Dirac delta with respect to $x$ evaluated at the point $g(x)$, i.e. $\delta'(g(x)) = \left.\frac{\mathrm{d}\delta}{\mathrm{d}x}\right|_{g(x)}$;
the derivative of the Dirac delta composed with $g$ with respect to $x$, i.e. $\delta'(g(x)) = \frac{\mathrm{d}}{\mathrm{d}x}\delta(g(x)) = \left.\frac{\mathrm{d}\delta}{\mathrm{d}x}\right|_{g(x)} \frac{\mathrm{d}g}{\mathrm{d}x}$ by the chain rule.
In your derivation, you were using the first definition; in that case, you should have had : $$ \begin{array}{rcl} \displaystyle \int_a^b \delta'(g(x))f(x) \,\mathrm{d}x &=& \displaystyle \int_a^b \frac{\mathrm{d}\delta(g(x))}{\mathrm{d}g}f(x) \,\mathrm{d}x \\ &=& \displaystyle \int_{g(a)}^{g(b)} \frac{\mathrm{d}\delta(y)}{\mathrm{d}y}f(g^{-1}(y)) \,\frac{\mathrm{d}y}{g'(g^{-1}(y))} \quad\mathrm{where}\quad y=g(x) \\ &=& \displaystyle \left.\delta(y)\frac{f(g^{-1}(y))}{g'(g^{-1}(y))}\right|_{y=g(a)}^{y=g(b)} - \int_{g(a)}^{g(b)} \delta(y)\frac{\mathrm{d}}{\mathrm{d}y}\left(\frac{f(g^{-1}(y))}{g'(g^{-1}(y))}\right) \,\mathrm{d}y \\ \end{array} $$ The main structure resembles your expression, however, we see that you forgot the Jacobian of the transformation $x \leftrightarrow y=g(x)$. Moreover, the boundary term doesn't vanish automatically; indeed, it is undefined if $g(a)=0$ or $g(b)=0$ because of the Dirac delta and the associated factor can diverge depending on the considered function $g$.
Instead, if you consider the more common definition n${}^\circ$2, you would have had : $$ \begin{array}{rcl} \displaystyle \int_a^b \delta'(g(x))f(x) \,\mathrm{d}x &=& \displaystyle \int_a^b \left.\frac{\mathrm{d}\delta}{\mathrm{d}x}\right|_{g(x)}f(x) \,\mathrm{d}x \\ &=& \displaystyle \left.\delta(g(x))f(x)\right|_{x=a}^{x=b} - \int_a^b \delta((g(x))f'(x) \,\mathrm{d}x \\ &=& \displaystyle \left.\delta(g(x))f(x)\right|_{x=a}^{x=b} - \sum_k\int_a^b \frac{\delta(x-x_k)}{|g'(x_k)|}f'(x) \,\mathrm{d}x \\ &=& \displaystyle \delta(g(b))f(b)-\delta(g(a))f(a) - \sum_k \frac{f'(x_k)}{|g'(x_k)|} 1_{(a,b)}(x_k) \\ \end{array} $$ where $x_k$ denotes the $k^\mathrm{th}$ zero of $g(x)$ and $1_{(a,b)}(x)$ is the indicator function of the interval $(a,b)$. Of course, this heavy expression will be simplified when you consider the "standard" domain of integration $(a,b) = \mathbb{R}$.
Finally, the third definition will produce an expression similar to the previous case by replacing $f(x)$ by $f(x)g'(x)$.