Delta Method corollary Consider the Delta Method as stated in van der Vaart Theorem 3.1 at page 26 (you can find the page here https://books.google.co.uk/books?id=UEuQEM5RjWgC&pg=PA32&lpg=PA32&dq=van+der+vaart+theorem+3.1+because+the+sequence+converges+in+distribution&source=bl&ots=mnRJLD8XLC&sig=inIMmSPvWDfrPc6r4U7dnuQ_3OM&hl=it&sa=X&ved=0ahUKEwj2sYfJob3JAhUKox4KHZNcBTUQ6AEILTAC#v=onepage&q=van%20der%20vaart%20theorem%203.1%20because%20the%20sequence%20converges%20in%20distribution&f=false). 
Notice that among the sufficient conditions it just requires the function $\phi()$ to be differentiable at $\theta$.
I have understood the proof but I don't know how to show that when $T$ is $N(0,\Sigma)$ then $\phi_{\theta}'(T)$ is $N(0,(\varphi'_{\theta})^T\Sigma (\varphi'_{\theta}))$. 
I have found this proof 
https://en.wikipedia.org/wiki/Delta_method
but it assumes $\phi()$ differentiable in the entire domain to apply the mean value theorem. 
Do you have suggestions on how to proceed?

If $T$ follows a Gaussian distribution on $\mathbb{R}^{n}$ with mean $0$ and covariance matrix $\Sigma$, denoted by $T \, \sim \, \mathcal{N}(0,\Sigma)$, and if $A$ is a $n \times n$ real matrix, we have :
$$ AT \, \sim \, \mathcal{N}(0,A \Sigma A^{\top}) $$
If $f \, : \, \mathbb{R}^{n} \, \rightarrow \, \mathbb{R}^{n}$ is differentiable at $\theta$, the differential of $f$ at $\theta$, denoted by $\mathrm{D}_{\theta}f$, is a linear map from $\mathbb{R}^{n}$ to itself. Let $\mathrm{J}(\theta)$ the jacobian matrix of $f$ at $\theta$. Therefore, for all $x$, $\mathrm{D}_{\theta}f \cdot x = \mathrm{J}(\theta)x$. 
$$ \mathrm{D}_{\theta}f \cdot T \, \sim \, \mathcal{N}( 0 , \mathrm{J}(\theta) \, \Sigma \, \mathrm{J}(\theta)^{\top} ) $$
