On the wikipedia page for the Dirac delta function derivative ( https://en.wikipedia.org/wiki/Dirac_delta_function#Distributional_derivatives currently by citation 50), It says that $$\delta'(-x) = -\delta'(x)$$
I'm not sure how this is possible. Because by definition of the derivative of the Dirac delta function: $$\int_{-\infty}^\infty\delta'(x)f(t)dt = -f'(x)$$
Meaning that $$\int_{-\infty}^\infty\delta'(-x)f(t)dt = -f'(-x)$$
But if $$\delta'(-x) = -\delta'(x)$$ is a true statement, then that would mean that $$-f'(-x) = f'(x)$$
Which is clearly not true for all functions.
You are making a confusion between $\delta(x) = \delta_0(x)$, which is the Dirac delta distribution (centered at $0$) $\delta = \delta_0$ seen as a "function" of the $x$ variable, and $\delta_x$, which is the Dirac delta centered at the point $x$.
They are defined as measures by the formulas for any nice function $\varphi$ $$ \langle \delta,\varphi\rangle = \int_{\Bbb R} \varphi(y)\,\delta(\mathrm d y) = \int_{\Bbb R} \varphi(y)\,\delta_0(\mathrm d y) = \varphi(0) \\ \langle \delta_x,\varphi\rangle = \int_{\Bbb R} \varphi(y)\,\delta_x(\mathrm d y) = \varphi(x). $$ The derivatives of these measures are no longer measures but distributions, so one should not use the notation $\int f\,\varphi$ but only $\langle f,\varphi\rangle$ (but you can think of them as being the same object) and then $$ \langle \delta'(-x),\varphi\rangle = \langle \delta',\varphi(-x)\rangle = -\langle \delta,(\varphi(-x))'\rangle \\ = \langle \delta,\varphi'(-x)\rangle = \varphi'(0) = \langle \delta,\varphi'\rangle = -\langle \delta',\varphi\rangle $$ so $\delta'(-x) = -\delta'(x)$. As you can see, it is due to the fact that $\varphi'(0) = \varphi'(-0)$. Of course this is no longer valid for $\delta_x$.