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?
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.