<p>The <a href="https://mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/dfi/t#p:slc" rel="nofollow noreferrer">implicit function theorem</a> is handy here.</p>
<p>Rewrite the equilibrium condition as
<span class="math-container">\begin{equation}
F(p,t):=S(p)-D(p+t) = 0
\end{equation}</span>
Using the implicit function theorem, we have
<span class="math-container">\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}</span>
as desired.</p>

CC BY-SA 4.0
2024-10-21T15:10:05.787Z
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 $d S(p) / dt = S' \cdot dp/dt$?
dchao
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2024-10-21T15:13:47.763Z
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 $dp/dt$.
dchao
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2024-10-21T15:16:28.900Z
And back to my question, in my approach, $D'(p+t)=D'(p)$ correct, right?
dchao
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2024-10-21T15:59:40.790Z
@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.
Herr K.
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2024-10-21T16:04:01.963Z
@ And in my first comment, do we treat $p$ and $t$ independent when we deal with the partial derivatives like $\partial F / \partial t$?
dchao
CC BY-SA 4.0
2024-10-21T16:12:57.087Z
Thank you very much. Yes. That is what I mean. If so, the statement $D'(p+t) \frac{d (p+t)}{d t}  \implies D'(p)(\frac{d p}{d t}+1)$ then is NOT correct?
dchao
CC BY-SA 4.0
2024-10-21T19:09:26.323Z
@dchao: Yes to treat $p$ and $t$ as independent. The issue with your derivation is the abuse of notation: $p$ stands both for the argument of $D$ and $S$, and for a particular value of it. To avoid equivocation, let $p$ denote the argument and $p_0$ be the price at which the market equilibrates. Then, 
$$
\frac{\partial D(p)}{\partial p}\frac{\partial p}{\partial t}\Bigg\vert_{p=p_0+t} 
= \frac{\partial D(p)}{\partial p}\Bigg\vert_{p=p_0+t}\frac{\partial (p_0+t)}{\partial t}
= D'(p_0+t)\left(\frac{\partial p}{\partial t}+1\right)
$$
Herr K.
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2024-10-22T04:48:03.010Z
Thanks Herr K.! I am fully appreciated of your answer!
dchao
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2024-10-21T13:45:19.650Z
2024-10-21T13:45:19.650Z
Herr K.
<p>Suppose we levy a specific tax <span class="math-container">$t$</span>, then, at the new equilibrium, we have
<span class="math-container">$$
D(p+t)=S(p),
$$</span>
where <span class="math-container">$D$</span> is the demand function and <span class="math-container">$S$</span> is the supply function. And this makes equilibrium price <span class="math-container">$p$</span> is a function of <span class="math-container">$t$</span>. The incidence formula
<span class="math-container">$$
\frac{dp}{dt} = \frac{D'}{S'-D'},
$$</span>
where <span class="math-container">$D'=dD(p)/dp$</span> and <span class="math-container">$S'=d S(p)/dp$</span>, can be obtained through <a href="https://scholar.harvard.edu/files/stantcheva/files/lecture3.pdf" rel="nofollow noreferrer">https://scholar.harvard.edu/files/stantcheva/files/lecture3.pdf</a>.</p>
<p>However, some texts state that we can also obtain this formula through implicit differentiation with respect to <span class="math-container">$t$</span>. <strong>My question is how we can do this.</strong></p>
<p>This is my approach:
<span class="math-container">\begin{aligned}
&amp;D(p+t)=S(p) \\
\implies &amp;\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 &amp;D'(p+t)(\frac{d p}{d t}+1)=S'(p) \frac{d p}{d t}
\end{aligned}</span>
It seems that only when <span class="math-container">$D'(p+t) = D'(p)$</span>, I can obtain the same tax incidence formula. <strong>But is <span class="math-container">$D'(p+t) = D'(p)$</span> TRUE?</strong></p>
<p>p.s. I know <a href="https://economics.stackexchange.com/questions/16229/is-there-an-intuitive-explanation-for-the-tax-incidence-formula-from-elasticity">Is there an intuitive explanation for the tax incidence formula from elasticity?</a> provides a good derivation way but it is still not through implicit differentiation.</p>

CC BY-SA 4.0
2024-10-21T13:17:30.053Z
Ah...I think it is certain that $\frac{d D(p+t)}{d (p+t)} = \frac{d D(u)}{d u}$...
dchao
<microeconomics>
Tax incidence formula derivation through implicit differentiation
dchao
# Tax incidence formula derivation through implicit differentiation

(Asked by: username_0 on 2024-10-21)

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?](https://economics.stackexchange.com/questions/16229/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.

**Tags:** microeconomics
> **Comments:**
> **username_0**: Ah...I think it is certain that $\frac{d D(p+t)}{d (p+t)} = \frac{d D(u)}{d u}$...  

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## 1 answers

### Answer by username_1 on 2024-10-21. Score: 3 | Accepted answer

The [implicit function theorem](https://mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/dfi/t#p:slc) 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.

> **Comments:**
> **username_0**: 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 $d S(p) / dt = S' \cdot dp/dt$?  
> **username_0**: 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 $dp/dt$.  
> **username_0**: And back to my question, in my approach, $D'(p+t)=D'(p)$ correct, right?  
> **username_1**: @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.  
> **username_0**: @ And in my first comment, do we treat $p$ and $t$ independent when we deal with the partial derivatives like $\partial F / \partial t$?  
> **username_0**: Thank you very much. Yes. That is what I mean. If so, the statement $D'(p+t) \frac{d (p+t)}{d t}  \implies D'(p)(\frac{d p}{d t}+1)$ then is NOT correct?  
> **username_1**: @dchao: Yes to treat $p$ and $t$ as independent. The issue with your derivation is the abuse of notation: $p$ stands both for the argument of $D$ and $S$, and for a particular value of it. To avoid equivocation, let $p$ denote the argument and $p_0$ be the price at which the market equilibrates. Then,  $$ \frac{\partial D(p)}{\partial p}\frac{\partial p}{\partial t}\Bigg\vert_{p=p_0+t}  = \frac{\partial D(p)}{\partial p}\Bigg\vert_{p=p_0+t}\frac{\partial (p_0+t)}{\partial t} = D'(p_0+t)\left(\frac{\partial p}{\partial t}+1\right) $$  
> **username_0**: Thanks Herr K.! I am fully appreciated of your answer!  

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