let $x_1,...,x_n$, $n\geq 3$ be a sample form the distribution $F$. We want to estimate $\frac{dF^{-1}(p)}{dp}$, replacing the distribution function $F$ by empirical distribution function $F_n$, and using a difference operator in place of differential operator.
We can write $F_n(x)=\frac{1}{n}\sum_{i=1}^n I(x\geq x_k)$, and, $F_n^{-1}(p)=\inf\{x:F_n(x)\geq p\}.$ Let $i$ be such that $\frac{i-1}{n}\leq p\leq \frac{i}{n}$. Then required estimation
\begin{align}\widehat{\frac{d}{dp}F^{-1}(p)}&=\Delta F_n^{-1}(p)\\ &=\Delta \inf\{x:F_n(x)\geq p\}\\ \end{align} Consider $F_n(x)\geq p$ for some $x$. The set of such $x$ form an interval $[x_{(i-m)},x_{(i+m)}]$, where $m\in\mathbb{Z^+}$. Then
$$\widehat{\frac{d}{dp}F^{-1}(p)}=\Delta [x_{(i-m)},x_{(i+m)}] $$ $$\widehat{\frac{d}{dp}F^{-1}(p)}=n\frac{x_{(i+m)}-x_{(i-m)}}{2m}$$
Is it correct way of doing? Kindly help me. Thanks beforehand!