$\psi(x,t)=Ae^{i(kx-\omega t)}+ Be^{-i(kx+\omega t)}$
If I want to find $\nabla\psi$, am I correct just to take the partial derivative of $\psi$ with respect to $x$, $\partial_x\psi$?
Does $\nabla\psi = Axe^{i(kx-\omega t)}+Bxe^{-i(kx+\omega t)}$?
$\psi(x,t)=Ae^{i(kx-\omega t)}+ Be^{-i(kx+\omega t)}$
If I want to find $\nabla\psi$, am I correct just to take the partial derivative of $\psi$ with respect to $x$, $\partial_x\psi$?
Does $\nabla\psi = Axe^{i(kx-\omega t)}+Bxe^{-i(kx+\omega t)}$?