<p>I will denote the demand function by <span class="math-container">$Q(p)$</span> and the inverse demand function by <span class="math-container">$P(q)$</span>. Then
<span class="math-container">$$
\forall q: Q(P(q)) = q
$$</span>
so for any <span class="math-container">$h &gt; 0$</span> and <span class="math-container">$q$</span> we have
<span class="math-container">$$
\begin{align*}
p &amp; := P(q) \\
p_h &amp; := P(q+h) \\
q &amp; = Q(p) \\
q_h &amp; := Q(p_h) = q+h
\end{align*}
$$</span>
From the definition of derivatives
<span class="math-container">$$
\frac{\text{d} P(q)}{\text{d} q} := \lim_{h \to 0} \frac{p_h - p}{q_h - q}.
$$</span>
If this is a non zero real number then indeed
<span class="math-container">$$
\frac{1}{\lim_{h \to 0} \frac{p_h - p}{q_h - q}} = \lim_{h \to 0} \frac{q_h - q}{p_h - p} = \frac{\text{d} Q(p)}{\text{d} p}.
$$</span></p>
<hr />
<p>There are some technical considerations. We did not prove that
<span class="math-container">$$
\lim_{h \to 0} p_h - p = 0.
$$</span>
This is true if <span class="math-container">$P$</span> is continuous.</p>
<p>We also assumed that at <span class="math-container">$q$</span> and <span class="math-container">$q+h$</span> both functions are single valued. If there is a price <span class="math-container">$p$</span> where <span class="math-container">$Q(p) = 0$</span> then <span class="math-container">$\frac{\text{d} P(q)}{\text{d} q}$</span> does not exist.</p>

CC BY-SA 4.0
2020-06-26T08:46:56.460Z
2020-06-26T08:46:56.460Z
Giskard
<p>Yes, for the standard case of a strictly decreasing demand function <span class="math-container">$Q(p)$</span> and price-elasticity of demand <span class="math-container">$\epsilon_p(Q)=Q'(p)\frac{p}{Q(p)}$</span> the inverse demand function <span class="math-container">$p(Q)$</span> exists and by the inverse function theorem <span class="math-container">$p'(Q)=\frac{1}{Q'(p)}$</span>. This gives <span class="math-container">$p'(Q)=\frac{p(Q)}{\epsilon_p(Q)Q}$</span> wherever the derivatives exist.</p>

CC BY-SA 4.0
2020-06-26T08:57:11.473Z
2020-06-26T15:30:30.100Z
2020-06-26T15:30:30.100Z
VARulle
<p>I was wondering of my thinking here was right:</p>
<p>Given <span class="math-container">$$ e=\frac{dQ}{dp}*\frac{p}{Q}, $$</span> where
<span class="math-container">$ e $</span> is elasticity, <span class="math-container">$ dQ/dp $</span> is first derivative of demand function, <span class="math-container">$p$</span> is price and <span class="math-container">$Q$</span> is quantity.</p>
<p>With this expression for <span class="math-container">$e$</span>, could you then state that:</p>
<p><span class="math-container">$$ \frac{dp}{dQ} = \frac{p}{(e*Q)} $$</span></p>
<p>where <span class="math-container">$ dp/dQ $</span> expresses first derivative of inverse demand.</p>
<p>Is this correct?</p>

<microeconomics><elasticity><demand>
elasticity from inverse demand
M H
# elasticity from inverse demand

(Asked by: username_0 on 2020-06-26)

I was wondering of my thinking here was right:

Given $$ e=\frac{dQ}{dp}\*\frac{p}{Q}, $$ where
$ e $ is elasticity, $ dQ/dp $ is first derivative of demand function, $p$ is price and $Q$ is quantity.

With this expression for $e$, could you then state that:

$$ \frac{dp}{dQ} = \frac{p}{(e\*Q)} $$

where $ dp/dQ $ expresses first derivative of inverse demand.

Is this correct?

**Tags:** microeconomics, elasticity, demand

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## 2 answers

### Answer by username_1 on 2020-06-26. Score: 4

I will denote the demand function by $Q(p)$ and the inverse demand function by $P(q)$. Then
$$
\forall q: Q(P(q)) = q
$$
so for any $h > 0$ and $q$ we have
$$
\begin{align\*}
p & := P(q) \\
p\_h & := P(q+h) \\
q & = Q(p) \\
q\_h & := Q(p\_h) = q+h
\end{align\*}
$$
From the definition of derivatives
$$
\frac{\text{d} P(q)}{\text{d} q} := \lim\_{h \to 0} \frac{p\_h - p}{q\_h - q}.
$$
If this is a non zero real number then indeed
$$
\frac{1}{\lim\_{h \to 0} \frac{p\_h - p}{q\_h - q}} = \lim\_{h \to 0} \frac{q\_h - q}{p\_h - p} = \frac{\text{d} Q(p)}{\text{d} p}.
$$

---

There are some technical considerations. We did not prove that
$$
\lim\_{h \to 0} p\_h - p = 0.
$$
This is true if $P$ is continuous.

We also assumed that at $q$ and $q+h$ both functions are single valued. If there is a price $p$ where $Q(p) = 0$ then $\frac{\text{d} P(q)}{\text{d} q}$ does not exist.


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### Answer by username_2 on 2020-06-26. Score: 3

Yes, for the standard case of a strictly decreasing demand function $Q(p)$ and price-elasticity of demand $\epsilon\_p(Q)=Q'(p)\frac{p}{Q(p)}$ the inverse demand function $p(Q)$ exists and by the inverse function theorem $p'(Q)=\frac{1}{Q'(p)}$. This gives $p'(Q)=\frac{p(Q)}{\epsilon\_p(Q)Q}$ wherever the derivatives exist.


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