How to show for a distribution $T$ and a test function $\varphi,~~T'[\varphi]\equiv -T[\varphi']\;?$ 
For a generalized
  function $T,$ we define
  $$T'[\varphi] ~≡~ −T[φ']~~~~~~\forall φ ∈ \mathcal D(Ω).$$
where $\mathcal D(\Omega)$ denotes the test function space.

I'm not getting how they deduced this relation. Can anyone tell me how to prove the relation?
 A: As has been mentioned in some comments, this is the definition of distributional derivative. In general, if $\alpha \in \mathbb{N}^n$ is a multi-index, the distributional derivative of order $\alpha$, of a distribution $T \in \mathcal{D}'(\Omega)$ is defined by placing
$(\star)$ $\displaystyle \langle D^\alpha T, \varphi \rangle =(-1)^{|\alpha|} \langle T, D^\alpha \varphi \rangle$ $\forall \varphi \in \mathcal{D}(\Omega)$
It is still a distribution, so $D^{\alpha}T \in \mathcal{D}'(\Omega)$ and it is essentially a generalization of the rule of integration by parts in $\mathbb{R}^n$.
Now, every $f \in L_{loc}^1(\Omega)$ determines a distribution $T_f$ defined by
$\displaystyle T_{f}(\varphi):= \int_{\Omega} f \varphi dx $ $\forall \varphi \in \mathcal{D}(\Omega)$
Then, for example, if $f \in C^1(\mathbb{R})$ we have $f' \in C(\mathbb{R})$ and  it is locally integrable, then we can define the distribution
$\displaystyle T_{f'}(\varphi)=\int_{\mathbb{R}} f'(x) \varphi(x) dx = f(x)\varphi(x) \vert_{|x|=\infty} - \int_{\mathbb{R}} f(x) \varphi'(x) dx = -T_{f}(\varphi')$
since $\varphi$ vanishes at infinity. Now $(\star)$ is the generalization of these facts in the distribution space.
