I was wondering about this:
Can we state that $\delta'(x)$ has the following property, specular to the Dirac Delta usual property?
$$\delta'(f(x)) = \sum_{\text{roots}} \frac{\delta'(x-x_i)}{|f'(x)|_{x = x_i}}$$
I am asking this in virtue of this problem:
$$\int_{-\infty}^{+\infty} \delta'(x^2-1) \phi(x)\ dx$$
In which I shall transform the Dirac Delta derivative, so I was wondering it if were possible to use
$$\delta'(x^2-1) = \frac{d}{dx} \delta(x^2-1) = \frac{d}{dx}\left(\frac{1}{2}(\delta(x+1) + \delta(x-1)\right) = \frac{1}{2}\left(\delta'(x+1) + \delta'(x-1)\right)$$
Also from this I would be fine because I'd thence use
$$\delta'(x\pm 1)\phi(x) = \delta(x\pm 1)\phi'(x) = \phi'(\pm 1)$$
We can proceed as follows:
$$\begin{align} \int_I \delta'(f(x)) \, \phi(x) \, dx & = \left. -\frac{d}{dy} \left( \phi(f^{-1}(y)) \, \frac{1}{|f'(f^{-1}(y))|} \right) \right|_{y=0} \\ & = \left. - \left( \phi'(f^{-1}(y)) \frac{1}{f'(f^{-1}(y))} \cdot \frac{1}{|f'(f^{-1}(y))|} + \phi(f^{-1}(y)) \cdot \frac{-f''(f^{-1}(y))}{|f'(f^{-1}(y))|^3} \right) \right|_{y=0} \\ & = - \left( \phi'(x_0) \frac{1}{f'(x_0)} \cdot \frac{1}{|f'(x_0)|} + \phi(x_0) \cdot \frac{-f''(x_0)}{|f'(x_0)|^3} \right) \\ & = \frac{f''(x_0)}{|f'(x_0)|^3} \phi(x_0) - \frac{f'(x_0)}{|f'(x_0)|^3} \phi'(x_0) \end{align}$$
Therefore our final formula becomes $$\delta'(f(x)) = \sum_n \frac{f''(x_n) \delta(x-x_n) + f'(x_n) \delta'(x-x_n)}{|f'(x_n)|^3}$$