Derivative of a delta function reference

Question

This question has been asked many times. I've read that $\dfrac{d(\delta(f(x)))}{dx}=-f^{'}(x)$. For instance, here, the second answer. I am coming from the context of probability. In particular, I am considering $\lim_{\sigma \rightarrow 0}N(f(x),\sigma)$, and that is how $\delta((f(x)))$ shows up. I would like a reference to a math book or something rigoruous that shows the result presented above, namely that $\dfrac{d(\delta(f(x)))}{dx}=-f^{'}(x)$. I've done some google search, so far I've seen many proofs that are not truly rigoruous, including this; in other posts they say that is so by definition, like here or here, is that so?. That is why I think a book would be ideal, but not necessary nonetheless. Any help would be appreciated.

Answer 1

So, now that it has become more clear, let answer more precisely.

Distributions: A distribution $g$ (from the theory of distribution of Laurent Schwartz) is defined as a linear form on the space of compactly supported functions $C^\infty_c$, and we write for every $\varphi∈C^\infty_c$ $$ g(\varphi) = \langle g,\varphi\rangle $$ Locally integrable functions can be seen as distributions by defining $\langle g,\varphi\rangle = ∫ g\varphi$ in this case. Some distributions are not functions, such as the Dirac delta defined by $$ \langle \delta_{a},\varphi\rangle := \varphi(a). $$ The derivative of a distribution $g$ is defined as the distribution $g'$ acting on smooth functions in the following way $$ \langle g',\varphi\rangle := -\langle g,\varphi'\rangle. $$ It is in particular compatible with the case when $g$ is a $C^1$ function by integrating by parts. Moreover, $\langle \delta'_{a},\varphi\rangle := -\varphi'(a).$

You problem: In your case, the normal law centered in $f(x)$, $N(f(x),\sigma)$ converges to the law with distributions $\delta_{f(x)}$. Now we want to compute the derivative with respect to $x$. Let $\varphi\in C^\infty_c$ be a test function. Then (by a theorem similar to the theorem of derivation under the integral sign) $$ \left\langle \frac{\mathrm{d}}{\mathrm{d}x}(\delta_{f(x)}),\varphi\right\rangle = \frac{\mathrm{d}}{\mathrm{d}x}\left\langle \delta_{f(x)},\varphi\right\rangle = \frac{\mathrm{d}}{\mathrm{d}x}\varphi(f(x)) = f'(x)\, \varphi'(f(x)) $$ and by the above definitions $$ \varphi'(f(x)) = \left\langle \delta_{f(x)},\varphi'\right\rangle = -\left\langle \delta_{f(x)}',\varphi\right\rangle. $$ We conclude that $$ \frac{\mathrm{d}}{\mathrm{d}x}(\delta_{f(x)}) = -f'(x)\,\delta_{f(x)}' $$