Fundamental property of Green's function is violated

Question

$$\langle x| D|x'\rangle=D_x\langle x|x'\rangle=D_x\,\delta(x-x')$$

$DD^{-1} = I$ , Where '$I$ ' represents the Identity

position representation of this equation is

$\langle x|D|x'\rangle \langle x|D^{-1}|x'\rangle=\langle x|I|x'\rangle=\delta(x-x')$

this is equivalent to

$\langle x|D|x'\rangle G(x,x')=\delta(x-x')$

Where $G(x,x')$ represents Green's function

This can be written using first equation as

$D_x\,\delta(x-x')G(x,x')=\delta(x-x')$

This shows fundamental property of green function that is

$D_x\,G(x,x')=\delta(x-x')$ is wrong ,

I don't Know where I am wrong, please clear my misconcept by spending your very precious little time

Answer 1

$$ \delta(x-x') = \langle x | x' \rangle = \langle x | I | x' \rangle = \langle x | D D^{-1} | x' \rangle = \int \langle x | D | x'' \rangle \langle x'' | D^{-1} | x' \rangle \, dx'' \\ = \int \langle x | D | x'' \rangle G(x'',x') \, dx'' = \int \langle x | x'' \rangle \partial_{x''} G(x'',x') \, dx'' \\ = \int \delta(x-x'') \partial_{x''} G(x'',x') \, dx'' = \partial_{x} G(x,x') $$