A demand function is defined as $q(p) = \dfrac{1}{p}$. The current price is $p=0.05$, with demand equal to $q(0.05) = 20$. What happens approximately with demand if the price is increased to $p=0.06$? And what if the price is decreased to $p=0.045$?
We solve this problem by the use of the property of the derivative
$$\Delta q = q(p +\Delta p) - q(p) \approx q'(p) \Delta p.$$
We need $q'(0.05)$:
$$
\begin{align}
q(p) &= \dfrac{1}{p} = p^{-1}\\
q'(p) &= -1 \cdot p^{-1-1} = -p^{-2} = \dfrac{-1}{p^2}\\
q'(0.05) &= \dfrac{-1}{(0.05)^2} = -400.
\end{align}
$$
If we increase the price to $p=0.06$, then $\Delta p = 0.06 - 0.05 = 0.01$. It holds that:
$$ \Delta q \approx q'(0.05) \cdot \Delta p = -400 \cdot 0.01 = -4.$$
Demand decreases by approximately 4 if the price increases by 0.01.
If we decrease the price to $p=0.045$, then $\Delta p = 0.045 - 0.05 = -0.005$. It holds that:
$$ \Delta q \approx q'(0.05) \cdot \Delta p = -400 \cdot -0.005 =2.$$
Demand increases by approximately 2 if the price decreases by 0.005.
Of course we can determine the change in $q$ also exactly in this easy example:
$$\begin{align}
q(0.05) &= \dfrac{1}{0.05}=20\\
q(0.06) &= \dfrac{1}{0.06} = 16\tfrac{2}{3}\\
q(0.045) &= \dfrac{1}{0.045} = 22\tfrac{2}{9}
\end{align}$$
If the price increases by $0.01$, then demand decreases by $20-16\tfrac{2}{3} = 3\tfrac{1}{3}$ and if the price decreases by $0.005$, then demand increases by $22\tfrac{2}{9} - 20 = 2\tfrac{2}{9}$.
We solve this problem by the use of the property of the derivative
$$\Delta q = q(p +\Delta p) - q(p) \approx q'(p) \Delta p.$$
We need $q'(0.05)$:
$$
\begin{align}
q(p) &= \dfrac{1}{p} = p^{-1}\\
q'(p) &= -1 \cdot p^{-1-1} = -p^{-2} = \dfrac{-1}{p^2}\\
q'(0.05) &= \dfrac{-1}{(0.05)^2} = -400.
\end{align}
$$
If we increase the price to $p=0.06$, then $\Delta p = 0.06 - 0.05 = 0.01$. It holds that:
$$ \Delta q \approx q'(0.05) \cdot \Delta p = -400 \cdot 0.01 = -4.$$
Demand decreases by approximately 4 if the price increases by 0.01.
If we decrease the price to $p=0.045$, then $\Delta p = 0.045 - 0.05 = -0.005$. It holds that:
$$ \Delta q \approx q'(0.05) \cdot \Delta p = -400 \cdot -0.005 =2.$$
Demand increases by approximately 2 if the price decreases by 0.005.
Of course we can determine the change in $q$ also exactly in this easy example:
$$\begin{align}
q(0.05) &= \dfrac{1}{0.05}=20\\
q(0.06) &= \dfrac{1}{0.06} = 16\tfrac{2}{3}\\
q(0.045) &= \dfrac{1}{0.045} = 22\tfrac{2}{9}
\end{align}$$
If the price increases by $0.01$, then demand decreases by $20-16\tfrac{2}{3} = 3\tfrac{1}{3}$ and if the price decreases by $0.005$, then demand increases by $22\tfrac{2}{9} - 20 = 2\tfrac{2}{9}$.