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?

By theorem 3.1 we know that if $\sqrt{n}(T_n-\theta)\rightsquigarrow T$, where $T\sim$ N$(\mu,\Sigma)$, then $\sqrt{n}(\phi(T_n)-\phi(\theta))\rightsquigarrow\phi_{\theta}'(T)$. Using the notation of Van Der Vaart we have $\phi_{\theta}'(T) = \phi'(\theta)T$. The distribution of this random variable is given by $\phi'(\theta)T\sim$ N$(\phi'(\theta)\mu,\phi'(\theta)\Sigma\phi'(\theta)^T)$ A proof of this last result can be found in corollary 2.5 of this syllabus (also written by Van der Vaart).
