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) $$ \frac{\partial L_1(x^*_1(p_1),x^*_2(p_2),p_1,p_2)}{\partial p_1} = \frac{\partial L_1}{\partial x_1}\frac{\partial x^*_1}{\partial p_1} $$


1 Answer 1


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: $$ \frac{\partial x_1}{\partial p_1} = - \frac{\alpha}{\alpha - 1} \frac{w}{(p_1)^2}. $$

  • $\begingroup$ 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 $\endgroup$
    – Three Diag
    Commented Feb 7 at 13:48

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