What is the first derivative of Dirac delta function?
If you imagine a Dirac delta impulse as the limit of a very narrow very high rectangular impulse with unit area centered at $t=0$, then it's clear that its derivative must be a positive impulse at $0^-$ (because that's where the original impulse goes from zero to a very large value), and a negative impulse at $0^+$ (where the impulse goes from a very large value back to zero).
Actually, the generalized derivative of a Dirac delta impulse $\delta(t)$, denoted by $\delta'(t)$, is a generalized function (distribution) with the following properties:
$$\begin{align}\int_{-\infty}^{\infty}\delta'(t)f(t)dt&amp;=-\int_{-\infty}^{\infty}\delta(t)f'(t)dt=-f'(0)\tag{1}\\\delta'(t)f(t)&amp;=f(0)\delta'(t)-f'(0)\delta(t)\tag{2}\\\int_{-\infty}^{\infty}\delta'(\tau)f(t-\tau)d\tau&amp;=f'(t)\qquad\qquad\textrm{(convolution)}\tag{3}\end{align}$$
Property $(1)$ is basically the definition of the derivative of a distribution. The right-most equality is of course a consequence of considering the special distribution $\delta(t)$.
Property $(3)$ means that convolution with the derivative of a Dirac impulse results in the derivative of the convolved function. I.e., the distribution $\delta'(t)$ is the impulse response of an ideal differentiator.
From property $(1)$ (with $f(t)=1$) it follows that
$$\int_{-\infty}^{\infty}\delta'(t)dt=0\tag{4}$$
Furthermore, from $(1)$ it also follows that $\delta'(t)$ is odd, because for every even function $f(t)$ whose derivative exists at $t=0$ we have
$$\int_{-\infty}^{\infty}\delta'(t)f(t)dt=0$$
since for even (and differentiable) $f(t)$, $f'(0)=0$ holds.