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Suppose we levy a specific tax $t$, then, at the new equilibrium, we have $$ D(p+t)=S(p), $$ where $D$ is the demand function and $S$ is the supply function. And this makes equilibrium price $p$ is a function of $t$. The incidence formula $$ \frac{dp}{dt} = \frac{D'}{S'-D'}, $$ where $D'=dD(p)/dp$ and $S'=d S(p)/dp$, can be obtained through https://scholar.harvard.edu/files/stantcheva/files/lecture3.pdf.

However, some texts state that we can also obtain this formula through implicit differentiation with respect to $t$. My question is how we can do this.

This is my approach: \begin{aligned} &D(p+t)=S(p) \\ \implies &\frac{d D(p+t)}{d (p+t)} \frac{d (p+t)}{d t} = \frac{d S(p)}{d p)} \frac{d p}{d t} \\ \implies &D'(p+t)(\frac{d p}{d t}+1)=S'(p) \frac{d p}{d t} \end{aligned} It seems that only when $D'(p+t) = D'(p)$, I can obtain the same tax incidence formula. But is $D'(p+t) = D'(p)$ TRUE?

p.s. I know Is there an intuitive explanation for the tax incidence formula from elasticity? provides a good derivation way but it is still not through implicit differentiation.

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  • $\begingroup$ Ah...I think it is certain that $\frac{d D(p+t)}{d (p+t)} = \frac{d D(u)}{d u}$... $\endgroup$
    – dchao
    Commented Oct 21 at 13:17

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The implicit function theorem is handy here.

Rewrite the equilibrium condition as \begin{equation} F(p,t):=S(p)-D(p+t) = 0 \end{equation} Using the implicit function theorem, we have \begin{equation} \frac{\mathrm dp}{\mathrm dt} = -\frac{\partial F/\partial t}{\partial F/\partial p} = -\frac{-D'}{S'-D'} = \frac{D'}{S'-D'} \;, \end{equation} as desired.

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  • $\begingroup$ Thanks. I have a question: is $\partial F / \partial t = 0 - D' \cdot 1$? If it is true, it seems that $p$ is independent of $t$ so that $d S(p) /d t = 0$? Why is not $d S(p) / dt = S' \cdot dp/dt$? $\endgroup$
    – dchao
    Commented Oct 21 at 15:10
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    $\begingroup$ But from another perspective, I derive the result. $0=dF = F_p dp + F_t dt$. That is, $0 = (S'-D')dp + (-D') dt$. Then, we get the desired $dp/dt$. $\endgroup$
    – dchao
    Commented Oct 21 at 15:13
  • $\begingroup$ And back to my question, in my approach, $D'(p+t)=D'(p)$ correct, right? $\endgroup$
    – dchao
    Commented Oct 21 at 15:16
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    $\begingroup$ @dchao: You have to careful with the prime notation of derivative. Let $x$ be the argument of the function $D$. Then, $$D'(p+t)=\frac{\mathrm dD(x)}{\mathrm dx}\Bigg\vert_{x=p+t}\quad\text{and}\quad D'(p)=\frac{\mathrm dD(x)}{\mathrm dx}\Bigg\vert_{x=p}.$$ The two are equal only if $t=0$. However, regardless of the value of $x$, $D'(x)$ is still same function defined as $\frac{\mathrm dD(x)}{\mathrm dx}$, which is probably what you meant by the equality of the derivatives. $\endgroup$
    – Herr K.
    Commented Oct 21 at 15:59
  • $\begingroup$ @ And in my first comment, do we treat $p$ and $t$ independent when we deal with the partial derivatives like $\partial F / \partial t$? $\endgroup$
    – dchao
    Commented Oct 21 at 16:04

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