Pulling out partial derivatives?

[Cogswell]

I'm reading through the book Quantum Mechanics (Second Edition) by David J. Griffiths and it got to the part about proving that if you normalise a wave function, it stays normalised (Page 13).

That part that I don't get is how they say:

$\dfrac{i \hbar}{2m} \left( \Psi^* \dfrac{\partial^2 \Psi}{\partial x^2} - \dfrac{\partial^2 \Psi^*}{\partial x^2} \Psi \right) = \dfrac{\partial}{\partial x} \left[\dfrac{i \hbar}{2m} \left( \Psi^* \dfrac{\partial \Psi}{\partial x} - \dfrac{\partial \Psi^*}{\partial x} \Psi \right) \right]$

How can they just pull out a partial operator like that?
Because if you expand it out again it would give you:

$\dfrac{i \hbar}{2m} \left( \dfrac{\partial}{\partial x} \left[ \Psi^* \dfrac{\partial \Psi}{\partial x}\right] - \dfrac{\partial}{\partial x} \left[ \dfrac{\partial \Psi^*}{\partial x} \Psi \right] \right)$

The operator will be applied to the wrong $\Psi$ and also won't you need to apply the product rule to is as well?

 

[wotanub]

and also won't you need to apply the product rule to is as well?

If you do just what you said, you'll see that the extra terms cancel and you'll be back at the original expression. Take out a scrap of paper.

 

[lugita15]

If you apply the product rule to the first term of $\dfrac{i \hbar}{2m} \left( \dfrac{\partial}{\partial x} \left[ \Psi^* \dfrac{\partial \Psi}{\partial x}\right] - \dfrac{\partial}{\partial x} \left[ \dfrac{\partial \Psi^*}{\partial x} \Psi \right] \right)$, you'll get a term $\dfrac{\partial \Psi}{\partial x}\dfrac{\partial \Psi^*}{\partial x}$, which cancels out with a term $- \dfrac{\partial \Psi}{\partial x}\dfrac{\partial \Psi^*}{\partial x}$ you'll get if you apply the product rule to the second term.

 

[Cogswell]

Haha right, thank you. I thought there was a special property of partial derivatives that I didn't know.