We have generalised coordinates $q_i$ and a Lagrangian $L(q_i,q_i',t)$. A new set of coordinates is introduced, namely $q_i=q_i(Q_j,t)$. [Note that throughout the question, the dash $'$ means time derivative.]
Using $ q_i' = \frac{\partial q_i}{\partial Q_j} Q_j' + \frac{\partial q_i}{\partial t} \space \space (**)$, it is easy to see that $$\frac{\partial q_i'}{\partial Q_j'}=\frac{\partial q_i}{\partial Q_j}$$ but I'm not one hundred percent clear as to how to go about showing $$\frac{\partial q_i'}{\partial Q_j}=\frac{d}{dt} \left( \frac{\partial q_i}{\partial Q_j} \right)$$
The solution goes about this line of working: $$\frac{\partial q_i'}{\partial Q_k} = \frac{\partial^2 q_i}{\partial Q_j \partial Q_k} + \frac{\partial^2 q_i}{\partial t \partial Q_k} = \frac{d}{dt} \left( \frac{\partial q_i}{\partial Q_k}(Q_j(t),t) \right) \space \space (*)$$
We have $ q_i' = \frac{\partial q_i}{\partial Q_j} Q_j' + \frac{\partial q_i}{\partial t} \space \space (**)$ from earlier in the question, so I'm guessing the above is simply an application of the chain rule to the formula for $q_i'$, differentiating w.r.t. $Q_k$. So my question is: where does the $Q_j'$ in $(**)$ disappear to after differentiation (it's not in $(*)$), and how do we pass from the middle of $(*)$ to the RHS ( I feel like I pretty much know how to do this, I'm most curious if the missing $Q_j'$ is perhaps a typo in the solution)?
Many thanks for any answers.
Normally $q_i$ if expressed with different coordinate system would be $q_i=q_i(Q_1, Q_2, \dots, t)$ $$\dot{q_i}=\sum_{k=0}^{n}\frac{\partial q_i}{\partial Q_k}\dot{Q_k}+\frac{\partial q_i}{\partial t}$$
let $j\leq n$ $$\frac{\partial\dot{q_i}}{\partial\dot{Q_j}}=\sum_{k=0}^{n}\frac{\partial q_i}{\partial Q_k}\frac{\partial \dot{Q_k}}{\partial Q_j}+0$$
$$\frac{\partial\dot{q_i}}{\partial\dot{Q_j}}=\frac{\partial q_i}{\partial Q_j}$$ Where dot {.} has been used to represent time derivative