User joshphysics has already answered OP's 1st question. Concerning OP's 2nd question, one derives

$$i\hbar \delta(x-x^{\prime})~=~i\hbar\langle x  | x^{\prime} \rangle ~=~\langle x | [\hat{x},\hat{p}] | x^{\prime} \rangle
~=~\langle x | \hat{x}\hat{p} | x' \rangle-\langle x | \hat{p} \hat{x} | x' \rangle~=~(x-x^{\prime})\langle x | \hat{p} | x^{\prime} \rangle
~=~-i\hbar(x-x^{\prime})\frac{\partial}{\partial x}\delta(x-x^{\prime}).$$

In other words

$$\delta(x-x^{\prime})~=~-(x-x^{\prime})\frac{\partial}{\partial x}\delta(x-x^{\prime}),$$

which also follows by differentiating the identity

$$ (x-x^{\prime})\delta(x-x^{\prime})~=~0 $$

wrt $x$.