How do you differentiate the unconditional quantile function for y wrt to P(x=1) where x is a Bernoulli random variable?

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Suppose $X$ is a Bernoulli random variable (0,1).

The marginal cdf of $Y$ can be written.

$F_Y(y)$ = $Pr(X=1)*F_{Y|X=1}(y)$ + $Pr(X=0)*F_{Y|X=0}(y)$ (Eq. 1)

Replace $y$ with $q_Y(\tau;p)$, the quantile function for $y$, indexed by $p=Pr(X=1)$

We now have an implicit function in $p$. Assume it is differentiable and differentiate it.

According to Firpo, Fortin, Lemieux, Unconditional Quantile Functions, Econometrica (2009)

$dq_Y(\tau;p)/dp$ = $[F_{Y|X=1}(y) - F_{Y|X=0}(y)] / f_Y(y)$

What confuses me is that this ignores the fact that $q_Y(\tau;p)$ is on both sides of Equation 1.

However, when I try to differentiate wrt both sides of Equation 1 I don't get a sensible expression.

Any help would be appreciated.

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That's not what that paper says; using their notation, the paper says

$$\frac{dq_\tau(p)}{dp}=\frac{P(Y>q_\tau|X=1)-P(Y>q_\tau|X=0)}{f_Y(q_\tau)}.$$

Recalling that $F_Y(q_\tau)=\tau$ by definition of a quantile, we can obtain this result via implicit differentiation ($\tau$ is fixed):

$$\tau=F_Y (q_\tau)=pF_{Y|X=1} (q_\tau|X=1)+(1-p)F_{Y|X=0} (q_\tau|X=0)\\ \implies \frac{\partial}{\partial p}:\quad 0=pf_{Y|X=1} (q_\tau|X=1)\frac{dq_\tau(p)}{dp}+F_{Y|X=1} (q_\tau|X=1)\\+(1-p)f_{Y|X=0} (q_\tau|X=0)\frac{dq_\tau(p)}{dp}-F_{Y|X=0} (q_\tau|X=0)\\ =f_Y (q_\tau)\frac{dq_\tau(p)}{dp}+F_{Y|X=1} (q_\tau|X=1)-F_{Y|X=0} (q_\tau|X=0)\\ =f_Y (q_\tau)\frac{dq_\tau(p)}{dp}+P(Y>q_\tau|X=0)-P(Y>q_\tau|X=1), $$ and the result follows.