We consider the demand function $X(p)=\dfrac{100}{(p-3)^2}$, $(p > 3)$. We determine by the use of elasticity the approximate percentage change in $X$ if $p$ increases by $5 \%$, given that $p=10$.
$$\begin{align}
\epsilon & = X'(p)\cdot \frac{p}{X(p)}\\
& =\frac{-100\cdot 2(p-3)} {(p-3)^4} \cdot \frac{p}{\frac{100}{(p-3)^2}}\\
& = \frac{-200}{(p-3)^3}\cdot \frac{(p-3)^2p}{100}\\
& = \frac{-2p}{p-3}.
\end{align}$$
Hence, at $p=10$: $\epsilon=\dfrac{-2 \cdot 10}{10-3}=\dfrac{-20}{7}$.
Therefore, $\% \Delta X \approx \epsilon ~ \%\Delta p = \frac{-20}{7}\cdot 5 =-\frac{100}{7}=-14\frac{2}{7}$.
$$\begin{align}
\epsilon & = X'(p)\cdot \frac{p}{X(p)}\\
& =\frac{-100\cdot 2(p-3)} {(p-3)^4} \cdot \frac{p}{\frac{100}{(p-3)^2}}\\
& = \frac{-200}{(p-3)^3}\cdot \frac{(p-3)^2p}{100}\\
& = \frac{-2p}{p-3}.
\end{align}$$
Hence, at $p=10$: $\epsilon=\dfrac{-2 \cdot 10}{10-3}=\dfrac{-20}{7}$.
Therefore, $\% \Delta X \approx \epsilon ~ \%\Delta p = \frac{-20}{7}\cdot 5 =-\frac{100}{7}=-14\frac{2}{7}$.