<p>First, there is a slight mistake in your first order condition as <span class="math-container">$x_2$</span> needs to be in the numerator and <span class="math-container">$x_1$</span> in the denominator.</p>
<p>The two first order conditions can be written as:
<span class="math-container">$$
\alpha p_2 x_2 - (1-\alpha) p_1 x_1 = 0,\\
p_1 x_1 + p_2 x_2 - w = 0.
$$</span>
Let's take the derivative of both with respect to <span class="math-container">$p_1$</span>.
<span class="math-container">$$
\alpha p_2 \frac{\partial x_2}{\partial p_1} - (1- \alpha) x_1 - (1-\alpha) p_1 \frac{\partial x_1}{\partial p_1} = 0\\
x_1 + p_1 \frac{\partial x_1}{\partial p_1} + p_2 \frac{\partial x_2}{\partial p_1} = 0
$$</span></p>
<p>Solve the second for <span class="math-container">$\frac{\partial x_2}{\partial p_1}$</span> and substitute in the first one to get:
<span class="math-container">$$
- \alpha x_1 - \alpha p_1 \frac{\partial x_1}{\partial p_1} - (1-\alpha) x_1 - (1- \alpha) p_1 \frac{\partial x_1}{p_1} = 0.
$$</span>
This gives:
<span class="math-container">$$
\frac{\partial x_1}{\partial p_1} = -\frac{x_1}{p_1}.
$$</span>
We also know that <span class="math-container">$x_1 = \frac{\alpha}{\alpha - 1}\frac{w}{p_1}$</span>, so:
<span class="math-container">$$
\frac{\partial x_1}{\partial p_1} = - \frac{\alpha}{\alpha - 1} \frac{w}{(p_1)^2}.
$$</span></p>

CC BY-SA 4.0
2024-02-07T13:48:48.400Z
This is great, but it I'm still a bit suspicious that we are using the implicit function itself, shouldn't we be always able to get the result without it? I guess when we don't have a closed form, which is when the theorem really matters, things will play differently
Three Diag
CC BY-SA 4.0
2024-02-07T10:33:48.097Z
2024-02-07T10:33:48.097Z
tdm
<p>I am trying to derive price changes
<span class="math-container">$$\frac{\partial x_1^*(p_1,p_2,w)}{\partial p_1}$$</span> for Cobb Douglas utility functions <span class="math-container">$$u(x_1,x_2)=x_1^\alpha x_2^{(1-\alpha)}$$</span> and I must use the implicit function theorem to do so.</p>
<p>My set of conditions from Lagrangian, simplified to remove the multiplier, are</p>
<p><span class="math-container">$$
L_1 = \frac{\alpha}{1-\alpha}\frac{x_1}{x_2} - \frac{p_1}{p_2} = 0
$$</span></p>
<p><span class="math-container">$$
L_2 = p_1x_1+p_2x_2 - w = 0
$$</span>.</p>
<p>Clearly, these two equations define two implicit function
which can be obtained explicitly solving the system as <span class="math-container">$$x_1^*(p_1,p_2,w) = \frac{\alpha}{1-\alpha}\frac{w}{p_1}$$</span>.
Their derivatives according to the Implicit Function Theorem are obtained also by solving the system
<span class="math-container">$$
\frac{\partial L_1}{\partial x_1} \frac{\partial x^*_1}{\partial p_1} + \frac{\partial L_1}{\partial x_2} \frac{\partial x^*_2}{\partial p_1} = - \frac{\partial L_1}{\partial p_1} 
$$</span></p>
<p><span class="math-container">$$
\frac{\partial L_2}{\partial x_1} \frac{\partial x^*_1}{\partial p_1} + \frac{\partial L_2}{\partial x_2} \frac{\partial x^*_2}{\partial p_1} = - \frac{\partial L_2}{\partial p_1} 
$$</span>
with
<span class="math-container">$
\frac{\partial x^*_1}{\partial p_1}$</span> and <span class="math-container">$\frac{\partial x^*_2}{\partial p_1}$</span> the unknowns.</p>
<p>I have two points of confusion which make me doubt whether I derived the right set of conditions.</p>
<p>First, I know that the final answer ought to contain the wealth <span class="math-container">$w$</span> but I don't see how that would be possible since any derivative of <span class="math-container">$L_2$</span> would cancel it out.</p>
<p>Second, I think the conditions contain the implicit functions themselves, therefore I would use the chain rule to derive the right hand side of the equation (taking for simplicity that I know that they only depend on their own prices)
<span class="math-container">$$
\frac{\partial L_1(x^*_1(p_1),x^*_2(p_2),p_1,p_2)}{\partial p_1} = \frac{\partial L_1}{\partial x_1}\frac{\partial x^*_1}{\partial p_1}
$$</span></p>

<utility><cobb-douglas>
Confusion on deriving Walrasian price changes using IFT
Three Diag
# Confusion on deriving Walrasian price changes using IFT

(Asked by: username_0 on 2024-02-06)

I am trying to derive price changes
$$\frac{\partial x\_1^\*(p\_1,p\_2,w)}{\partial p\_1}$$ for Cobb Douglas utility functions $$u(x\_1,x\_2)=x\_1^\alpha x\_2^{(1-\alpha)}$$ and I must use the implicit function theorem to do so.

My set of conditions from Lagrangian, simplified to remove the multiplier, are

$$
L\_1 = \frac{\alpha}{1-\alpha}\frac{x\_1}{x\_2} - \frac{p\_1}{p\_2} = 0
$$

$$
L\_2 = p\_1x\_1+p\_2x\_2 - w = 0
$$.

Clearly, these two equations define two implicit function
which can be obtained explicitly solving the system as $$x\_1^\*(p\_1,p\_2,w) = \frac{\alpha}{1-\alpha}\frac{w}{p\_1}$$.
Their derivatives according to the Implicit Function Theorem are obtained also by solving the system
$$
\frac{\partial L\_1}{\partial x\_1} \frac{\partial x^\*\_1}{\partial p\_1} + \frac{\partial L\_1}{\partial x\_2} \frac{\partial x^\*\_2}{\partial p\_1} = - \frac{\partial L\_1}{\partial p\_1}
$$

$$
\frac{\partial L\_2}{\partial x\_1} \frac{\partial x^\*\_1}{\partial p\_1} + \frac{\partial L\_2}{\partial x\_2} \frac{\partial x^\*\_2}{\partial p\_1} = - \frac{\partial L\_2}{\partial p\_1}
$$
with
$
\frac{\partial x^\*\_1}{\partial p\_1}$ and $\frac{\partial x^\*\_2}{\partial p\_1}$ the unknowns.

I have two points of confusion which make me doubt whether I derived the right set of conditions.

First, I know that the final answer ought to contain the wealth $w$ but I don't see how that would be possible since any derivative of $L\_2$ would cancel it out.

Second, I think the conditions contain the implicit functions themselves, therefore I would use the chain rule to derive the right hand side of the equation (taking for simplicity that I know that they only depend on their own prices)
$$
\frac{\partial L\_1(x^\*\_1(p\_1),x^\*\_2(p\_2),p\_1,p\_2)}{\partial p\_1} = \frac{\partial L\_1}{\partial x\_1}\frac{\partial x^\*\_1}{\partial p\_1}
$$

**Tags:** utility, cobb-douglas

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## 1 answers

### Answer by username_1 on 2024-02-07. Score: 4

First, there is a slight mistake in your first order condition as $x\_2$ needs to be in the numerator and $x\_1$ in the denominator.

The two first order conditions can be written as:
$$
\alpha p\_2 x\_2 - (1-\alpha) p\_1 x\_1 = 0,\\
p\_1 x\_1 + p\_2 x\_2 - w = 0.
$$
Let's take the derivative of both with respect to $p\_1$.
$$
\alpha p\_2 \frac{\partial x\_2}{\partial p\_1} - (1- \alpha) x\_1 - (1-\alpha) p\_1 \frac{\partial x\_1}{\partial p\_1} = 0\\
x\_1 + p\_1 \frac{\partial x\_1}{\partial p\_1} + p\_2 \frac{\partial x\_2}{\partial p\_1} = 0
$$

Solve the second for $\frac{\partial x\_2}{\partial p\_1}$ and substitute in the first one to get:
$$
- \alpha x\_1 - \alpha p\_1 \frac{\partial x\_1}{\partial p\_1} - (1-\alpha) x\_1 - (1- \alpha) p\_1 \frac{\partial x\_1}{p\_1} = 0.
$$
This gives:
$$
\frac{\partial x\_1}{\partial p\_1} = -\frac{x\_1}{p\_1}.
$$
We also know that $x\_1 = \frac{\alpha}{\alpha - 1}\frac{w}{p\_1}$, so:
$$
\frac{\partial x\_1}{\partial p\_1} = - \frac{\alpha}{\alpha - 1} \frac{w}{(p\_1)^2}.
$$

> **Comments:**
> **username_0**: This is great, but it I'm still a bit suspicious that we are using the implicit function itself, shouldn't we be always able to get the result without it? I guess when we don't have a closed form, which is when the theorem really matters, things will play differently  

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