Given: $${\partial \over \partial x} = \cos\phi{\partial\over \partial \rho} -{\sin\phi\over \rho}{\partial\over\partial \phi}$$
I was told to differentiate to get: $${\partial^2\over\partial x^2}=\cos^2\phi{\left({\partial^2\over\partial \rho^2}\right)}+{\sin^2\phi\over \rho}{\left(\partial\over\partial \rho\right)}+{\sin2\phi\over \rho^2}{\left(\partial\over\partial\phi\right)}-{\sin2\phi\over\rho}{\left(\partial^2\over\partial\rho\partial\phi\right)}+{\sin^2\over\rho^2}{\left(\partial^2\over\partial\phi^2\right)}$$ I understand how to get the first, fourth, and fifth terms, but a little confused with the second and third. For the third term, I came up with: $${\sin2\phi\over \rho^2}{\left(\partial\over\partial\phi\right)}=\left(-2\sin\phi\cos\phi\left(\partial\over\partial\phi\right)\right)\left({\partial\over\partial\rho}\left({1\over\rho}\right)\right)={2\sin\phi\cos\phi\over\rho^2}\left({\partial\over\partial\phi}\right)$$ Finally, for the second term: $${\sin^2\phi\over\rho}\left(\partial\over\partial\rho\right)={\left({-\sin\phi\over\rho}\left(\partial\over\partial\rho\right)\right)}{\left({\partial\over\partial\phi}\left(\cos\phi\right)\right)}$$ Can someone please explain how the product rule works with partial derivatives?
Edit: why isn't the second term $${2\sin^2\phi\over\rho}\left(\partial\over\partial\rho\right)$$ Since there are two terms like that added together?