Momentum operator in the position basis

Question

J.J Sakurai shows in the section of ' Momentum operator in the position basis' as

$P$$\lvert\alpha\rangle$=$\int dx^{'}\lvert\ x{'}\rangle\Bigl(-i{h\over 2\pi}$ $\partial\over\partial x{'}$$ \langle\ x{'}\rvert \alpha\rangle \Bigr)$

this gives

$\langle\ x{'}\rvert P\lvert\alpha\rangle$=$-i{h\over 2\pi}$ $\partial\over\partial x{'}$$ \langle\ x{'}\rvert \alpha\rangle $

I don't get this step can anyone please help me?

A little help would really be appreciated.

Answer 1

\begin{eqnarray} P | \alpha \rangle &=& -i\hbar \int{\rm d}y |y\rangle \frac{\partial}{\partial y}\langle y| \alpha\rangle \\ \langle x' | P | \alpha \rangle &=& -i\hbar \langle x'| \int{\rm d}y |y\rangle \frac{\partial}{\partial y}\langle y| \alpha\rangle \\ &=& -i\hbar \int{\rm d}y \color{blue}{\langle x'|y\rangle} \frac{\partial}{\partial y}\langle y| \alpha\rangle \\ &=& -i\hbar \int{\rm d}y \color{blue}{\delta(x' - y)}\frac{\partial}{\partial y}\langle x| \alpha\rangle \\ &=& -i\hbar \frac{\partial}{\partial x'}\langle x'| \alpha\rangle \end{eqnarray}