Partial Differential Equations – Demonstrating the Partial Derivative of Lagrangian

Question

I got the Lagrangian equation
$$L(q,\dot{q})=T-V=\frac{1}2m|(\dot{\textbf{q}}+\textbf{S}(\omega)\textbf{q})|^2-V(\textbf{Rq})$$

where S and R are 3×3 matrices. I have to show that

$$\frac{\partial L(\textbf{q},\dot{\textbf{q}})}{\partial \dot{\textbf{q}}}=m(\dot{\textbf{q}}+\textbf{S}(\omega)\textbf{q})$$
$$\frac{\partial L(\textbf{q},\dot{\textbf{q}})}{\partial \textbf{q}}=-m\textbf{S}(\dot{\textbf{q}}+\textbf{S}(\omega)\textbf{q})-\frac{\partial}{\partial \textbf{q}}V(\textbf{Rq})$$

I am told that it might be convenient to go into coordinates, i.e., work out the k'th Euler-Lagrange equation.

In the first partial derivative it seems that $-V(\textbf{Rq})$ disapears when we take the partial derivative with respect to $\dot{\textbf{q}}$ and i would have guessed that we had to use the chain rule for the squared part but it seems that isnt necessary?

For the second part I cant wrap my head around how you get $-m\textbf{S}(\dot{\textbf{q}}+\textbf{S}(\omega)\textbf{q})$

I also have to show that the Euler-Lagrange equations consequently becomes

$$m(\ddot{\textbf{q}}+2\textbf{S}(\omega)\dot{\textbf{q}}+\textbf{S}(\omega)\textbf{S}(\omega)\textbf{q}+\textbf{S}(\dot{\omega})\textbf{q})+\frac{\partial}{\partial \textbf{q}}V(\textbf{Rq})=0$$

Can anyone help me out here?

Thanks

EDIT
Guy3141 came with a super answer so the question is already partly solved, but..
When I doint the partial derivative for q I get

$$\frac{\partial L(\textbf{q},\dot{\textbf{q}})}{\partial \textbf{q}}=m\textbf{S}(\dot{\textbf{q}}+\textbf{S}(\omega)\textbf{q})-\frac{\partial}{\partial \textbf{q}}V(\textbf{Rq})$$

and I dont get the minus sign in front of the right hand side. Can anyone explain where the minus gets into the equation?

EDIT EDIT
I've been told to write everything in coordinates so that $q=(q_1,q_2,q_3)$ should do the trick and show why the minus sign is there, but I still don't see where the minus sign comes from 🙁 Anyone able to help here?

Thanks

Answer 1

How about we write $$|\dot{\mathbf{q}}+\mathbf{S}q|^2=\langle\dot{\mathbf{q}}+\mathbf{S}q,\dot{\mathbf{q}}+\mathbf{S}q \rangle$$ and use rule for differentiating an inner product.