Derivative of Dirac delta function Is the relation of the Dirac delta function correct?
$$
\frac{\partial}{\partial x''}\delta(x''-x')
= -\frac{\partial}{\partial x'}\delta(x'-x'').\tag{1}
$$
If it is, how to derive the above relation?  
 A: Define new coordinates $$x^{\pm}~:=~x^{\prime}\pm x^{\prime\prime}.\tag{A}$$
Then the chain rule yields
$$\left(\frac{\partial}{\partial x^{\prime}}+\frac{\partial}{\partial x^{\prime\prime}}\right)\delta(x^{\prime}-x^{\prime\prime})
~=~ 2\frac{\partial}{\partial x^+}\delta(x^-)~=~0,\tag{B}$$
which is OP's sought-for identity (1). 
A: If you meant $T(x,y) = \delta(x-y)$ the distribution on $\Bbb{R}^2$ defined by $<T,\phi> = \int_{-\infty}^\infty \phi(t,t)dt$ for $\phi \in C^\infty_c(\Bbb{R}^2)$
and $\partial_x T$ the distribution defined by 
$<\partial_x T,\phi> =-< T,\partial_x\phi>= -\int_{-\infty}^\infty \partial_x\phi(t,t)dt$ 
and
$<\partial_y T,\phi> =-< T,\partial_y\phi> =-\int_{-\infty}^\infty \partial_y\phi(t,t)dt$ 
then $\partial_x\phi(t,t)+\partial_y\phi(t,t) $ is the derivative of $t \mapsto  \phi(t,t)$ so that
$<\partial_x T+ \partial_y T,\phi> =- (\lim_{a\to \infty}\phi(a,a)+\phi(-a,-a) )= 0$ thus $\partial_x T+ \partial_y T=0$.
A: To prove
\begin{align}
\frac{\partial  \delta(x-x_0)}{\partial x}  = -\frac{\partial  \delta(x-x_0)}{\partial x'} . ~~~~~~~~~~~~~~~~~~~~ \text{(1)} 
\end{align}
Proof: We know that
\begin{align}
\int f(x) \frac{\partial \delta(x-x_0) }{\partial x}= - f'(x_0)   ~~~~~~~~~~~~~~~~~~~~~ \text{(2)}  
\end{align}
Now,
\begin{align}
& \int f(x) \frac{\partial \delta(x-x_0) }{\partial x_0}  dx  \\
= & \int f(x)  \left(  \frac{ \delta(x-(x_0+\epsilon) - \delta(x-x_0)    }{(x_0+\epsilon) - x_0} \right) dx   \\
= & \frac{  f(x_0 + \epsilon) -f(x_0 )  }{\epsilon} \\
= & f'(x_0),
\end{align}
which upon comparison with Eq. (2) immediately gives us Eq. (1).
