Wikipedia article under generalized forces says
Also we know that the generalized forces are defined as
How can I derive the first equation from the second for a monogenic system?
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TL;DR: Concerning OP's question (v3), it follows by definition without the use of eq. (2).
The first equation is the very definition of a monogenic force, i.e. that the force has a generalized (possibly velocity-dependent) potential.
The second equation is the definition of a generalized force.
Eq. (1) is a special case of eq. (2), but not vice-versa.
Example: A friction force is a generalized force that is not monogenic, cf. e.g. this Phys.SE post.
Qmechanic
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So the generalized force for a monogenic system which you call as monogenic force is just defined as having two terms and therefore can't be derived from the more general definition of the generalised force involving a summation? – Kashmiri Sep 19 '21 at 11:28
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$\uparrow$ Yes. – Qmechanic Sep 19 '21 at 11:29
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In the special case then is this correct $Q_{j}=\sum_{i=1}^{n} \mathbf{F}{i} \cdot \frac{\partial \mathbf{r}{i}}{\partial q_{j}}$=$-\frac{\partial \mathcal{V}}{\partial q_{i}}+\frac{d}{d t}\left(\frac{\partial \mathcal{V}}{\partial \dot{q}_{i}}\right)$ ? – Kashmiri Sep 21 '21 at 08:16
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$\uparrow$ Yes. – Qmechanic Sep 21 '21 at 08:19
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And we can't prove explicitly that the summation on the LHS equals the RHS for a monogenic case? – Kashmiri Sep 21 '21 at 08:22