Transformation Function - Position & Momentum Operators

[dengar768]

Homework Statement

I am currently studying for my quantum physics exam and I am trying to derive the Transformation function:
⟨x'│p' ⟩=Nexp{(ip' x')/ℏ}

Homework Equations

⟨x'│p' ⟩=Nexp{(ip' x')/ℏ}

The Attempt at a Solution

Now I get how to get to
p'⟨x'│p' ⟩=-iℏ d/dx' ⟨x'│p' ⟩

but i can't work out how to get both the p' and the x' into the expontential and I am not sure how to explain where the N comes from. Every example i find seems to jump from p'⟨x'│p' ⟩=-iℏ d/dx' ⟨x'│p' ⟩ straight to ⟨x'│p' ⟩=Nexp{(ip' x')/ℏ} without any explanation.
Would anybody be able to explain how to do this?
Thanks

 

[diazona]

Well, you have a differential equation,
$$p \langle x \vert p \rangle = -i\hbar \frac{\mathrm{d}}{\mathrm{d}x} \langle x \vert p \rangle$$
do you know how to solve it?

 

[dengar768]

No. Is this problem easier than I am making it?

 

[Cyosis]

Do you know how to solve $a f(x)=f'(x)$? This is exactly the same.