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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?

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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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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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