In some two player game in player $1$ and $2$ exert efforts $x_1$ and $x_2$ to win a prize with value $V$. The Nash Equilibrium is given by $argmax_{x_1}(\pi_1)$, and we have the following first order conditions:
$$\frac{\partial\pi_1}{\partial x_1}=g_1(x_1,x_2,V)-1=0$$
$$\frac{\partial\pi_2}{\partial x_2}=g_2(x_2,x_1,V)-1=0$$
I'm interested in how the value of the prize affects equilibrium level effort $\frac{\partial x_1^*}{\partial V}$, where $x_1^*$ denotes $1$'s equilibrium level effort. I know that $x_1^*$ is increasing in $V$ if we keep $x_2$ fixed, since
$$\frac{\partial g_1}{\partial V}>0 \hspace{0.2cm} ; \hspace{0.2cm} \frac{\partial g_1}{\partial x_1}<0$$
Which implies that when $V$ increases, $g_1$ increases, which needs to be offset by an increase in $x_1$ to satisfy the FOC. However, this also applies to $x_2$ and I know that player $1$ reduces $x_1$ by some rate $R$ after an increase in $x_2$. Hence we have i) $V$ increases $x_1^*$, ii) $V$ increases $x_2^*$ and, iii) $x_2^*$ reduces $x_1^*$. What condition would describe when the net effect of $V$ is positive or negative on $x_1^*$?
My attempt is as follows: $$\frac{\partial g_1/\partial V}{-\partial g_1/\partial x_1}$$ gives the extent to which $x_1$ increases following an increase in $V$ (as a ratio of $\epsilon$). Dividing this by the same for player $2$ gives the ratio denoted $A$: $$A= \frac{\partial g_1/\partial V}{-\partial g_1/\partial x_1}*\bigg(\frac{\partial g_2/\partial V}{-\partial g_2/\partial x_2}\bigg)^{-1}$$
Which gives ratio of the increase in $x_1^*$ to the increase in $x_2^*$. Hence, if
$$A>R$$
Then, the net effect of $V$ on $x_1^*$ is positive, and negative otherwise.
Let $x_2^*(x_1,V)$ be player $2$'s best response to $x_1$ with prize $V$. That is, $x_2^*$ satisfies
$$ g_2 (x_1,x_2^*,V)=1. $$
In equilibrium, player $1$'s effort $x_1^E$ satisfies
$$ g_1 (x_1^E,x_2^*(x_1^E,V),V)=1. $$
By the implicit function theorem, what we need to calculate is:
$$ \frac{\partial x_1^E}{\partial V} =- \frac{\frac{\partial g_1}{\partial x_2}\frac{\partial x_2^*}{\partial V} + \frac{\partial g_1}{\partial V}}{\frac{\partial g_1}{\partial x_1}+\frac{\partial g_1}{\partial x_2}\frac{\partial x_2^*}{\partial x_1}} .$$
This might be simplified somewhat by using the implicit function theorem to calculate the derivatives of $x_2^*$. The effect of $V$ on equilibrium effort now depends on the sign of $\frac{\partial x_1^E}{\partial V}$.