just a heads up - to solve this problem we must know about Lagrange's equation and expressions for generalised velocity, kinetic energy and generalised forces. It's not purely mathematics
Show that if kinetic energy does not depend on generalised coordinates, $q_j$ but only on time or $\dot{q_j},$ then $Q_j = \frac{\partial L}{\partial q_j},$ where $L$ is the Lagrangian.
I've worked out that
$$\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial T}{\partial \dot{q_j}} = \sum_i m_i \ddot{x_i}\frac{\partial x_i}{\partial q_j} + m_i \dot{x_i} \frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial x_i}{\partial q_j}$$
by $$\frac{\partial \dot{x_i}}{\partial \dot{q_j}} = \frac{\partial x_i}{\partial q_j}$$
where $T$ is kinetic energy. But by the chain rule, the second term in the first expression is
$$\frac{\partial T}{\partial q_j} = m_i\dot{x_i}\frac{\partial \dot{x_i}}{\partial q_j} = m_i\dot{x_i}\frac{\partial }{\partial q_j} \left (\ \sum_{k}\frac{\partial x_i}{\partial q_k}\dot{q_k} + \frac{\partial x_i}{\partial t} \right )$$
and the first term is $Q_j$. So the expression equals $Q_j + \frac{\partial T}{\partial q_j}$. Now Lagrange's equation also has a $-\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial V}{\partial \dot{q_j}}$ term and a $-\frac{\partial L}{\partial q_j}$ term on the LHS. Because we have that kinetic energy does not depend on coordinates, $Q_j -\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial V}{\partial \dot{q_j}} + \frac{\partial V}{\partial q_j} = 0$
I may well have messed up somewhere and I'm not sure if this method is even going to work. I don't think I really have the intuition to be able to solve this yet. Could anyone help?
The Lagrangian $\mathcal{L}$ is built up from the sum of the kinetik energy $T$ and the potential energy $V$ with its sign reversed, $i.e.$ $$\mathcal{L}=T-V\tag{*}$$
The equations of motion come from the minimum principle of the action. They are derived from: $$\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial\dot{q}}\right)=\frac{\partial \mathcal{L}}{\partial q}\tag{**}$$ where $q$ and $\dot{q}$ are the generalised coodinates.
If $T=T(\dot{q})$ and $V=V(q)$ then equation $(*)$ introduced in $(**)$ reads: $$\frac{d}{dt}\left(\frac{\partial T}{\partial\dot{q}}\right)=-\frac{\partial V}{\partial q}\tag{***}$$ The latter equation $(***)$ is the Newton second law, and it is clear that the generalised forces $Q$ are given by: $$Q=-\frac{\partial V}{\partial q}=\frac{\partial L}{\partial q}$$