Calculating total derivative of multivariable function

[halleffect]

Homework Statement

Here is the full problem statement (with equations): https://i.imgur.com/z2SWodC.png The answer that I'm trying to derive is at the bottom.

Relevant Equations

All equations are also included in the above image (https://i.imgur.com/z2SWodC.png).

This isn't a homework problem exactly but my attempt to derive a result given in a textbook for myself. Below is my attempt at a solution, typed up elsewhere with nice formatting so didn't want to redo it all. Direct image link here. Would greatly appreciate if anyone has any pointers.

 

[anuttarasammyak]

By $I_0$ and $V_0$ sum conditons, we know
i=2i_1-I_0=i(q_1)
v=2v_1-V_0=v(q_1)
They both are functions of variable $q_1$ only.
\frac{di}{dv}=\frac{\frac{di}{dq_1}} {\frac{dv}{dq_1}}=\frac{\frac{di_1}{dq_1}} {\frac{dv_1}{dq_1}}
we can calculate it.

 

[halleffect]

By $I_0$ and $V_0$ sum conditons, we know
i=2i_1-I_0=i(q_1)
v=2v_1-V_0=v(q_1)
They both are functions of variable $q_1$ only.
\frac{di}{dv}=\frac{\frac{di}{dq_1}} {\frac{dv}{dq_1}}=\frac{\frac{di_1}{dq_1}} {\frac{dv_1}{dq_1}}
we can calculate it.

Thanks very much for the reply; with that info:

\frac{di_1}{dq_1} = 2q_1+1
\frac{dv_1}{dq_1} = U_T(\frac{2q_1+1}{q_1})

And from this I get

\frac{di}{dv}=\frac{ \frac{di_1}{dq_1} }{ \frac{dv_1}{dq_1} }= \frac{q_1}{U_T}

I'm not sure if this is correct. For one thing, I'm not sure how I could incorporate $q_2$ into the answer like the given one (which is $\frac{2}{U_T}\frac{q_1 q_2}{q_1+q_2}$), although it does seem to make sense that it should be expressible through just one of the $q$'s. Is there a way to show this to be equal to the given answer? (Assuming I haven't made an error here)