Derivation of Dirac delta function Is there anyone could give me a hint how to find the distributional derivative of the delta function $\delta$? I don't know how to deal with the infinite point.

Upon request in the comments:
There is a large class of distributions which are given by integration against locally integrable functions. Specifically, given a locally integrable $f$ and a smooth compactly supported $g$, one can define $T_f(g)=\int_{-\infty}^\infty f(x) g(x) dx$. This leads to a common abuse of notation, where we write the same thing for distributions which are not given by integration against a locally integrable function. Thus we write things like $\delta(f)=\int_{-\infty}^\infty \delta(x) f(x) dx$, even though "$\delta(x)$" is meaningless by itself.
This abuse of notation turns out to be productive, because we can often define operations in distribution theory by passing to an approximating sequence, performing the operation on the approximating sequence (where it is defined the way we want) and then taking a limit. For example, if $T_{f_n} \to T$ and $f_n$ are smooth, then we can define $T'$ (the distributional derivative) to be $\lim_{n \to \infty} T_{f'_n}$. But with $T'_{f_n}$ we can really integrate by parts, which gives the formula $T'(g)=-T(g')$. Note that the result does not involve the approximation scheme, so we can just call this formula the "distributional derivative" and forget about approximating it.
The abuse of notation hints at trying such manipulations. Often people write them out without mentioning any approximation scheme. In this case the manipulations are called "formal", because they only pay attention to "form" without worrying about semantics. (This is the same use of the word as in "formal power series".)
Anyway, in the case of the Dirac delta, this procedure winds up telling you that $\delta'(f)=-f'(0)$ since by definition $\delta(f)=f(0)$.
