<p>In the second term of the intermediate calculation we must find the partial derivative of the entire second parameter of the $f$ function.  Hence
$$dU = \frac{\partial f}{\partial{q_1}}dq_1+\frac{\partial f}{\partial (\frac{y_0-p_1q_1}{p_2})}d(\frac{y_0-p_1q_1}{p_2})$$.  Obviously this is a substitution of $q_2$ into the original total differential derivation you had.  My point is that you tried to do the partial differential of $f$ with respect $q_2$ by assuming that this would equal $$\frac{dq_2}{dq_1}$$ when this may not always be the case.  For the final equation that you have may want to consider the chain rule.  The way you have it, we have defined $$U=f(q_1, q_2)$$ and $$q_2(q_1)=\frac{y_0}{p_2}-\frac{p_1}{p_2}$$  Then according to the chain rule you have $$\frac{dU}{dq_1}=\frac{\partial f}{\partial q_1}\frac{dq_1}{dq_1}+\frac{\partial f}{\partial q_2}\frac{dq_2}{dq_1}$$  Then we evaluate $$\frac{dq_1}{dq_1}=\frac{d}{q_1}q_1=1$$  Then we get $$\frac{dU}{dq_1}=\frac{\partial f}{\partial q_1}+\frac{\partial f}{\partial q_2}\frac{dq_2}{dq_1}$$  Then, $$\frac{dq_2}{dq_1}=-\frac{p_1}{p_2}$$ Finally we end up with $$\frac{dU}{dq_1}=f_1+f_2(-\frac{p_1}{p_2})$$</p>

CC BY-SA 3.0
2017-03-19T02:19:43.517Z
2017-03-19T02:19:43.517Z
user278039
<p>I am trying to fully understand the process of maximizing a utility function subject to a budget constraint while utilizing the Substitution Method (as opposed to the Lagrangian Method). I am following the work of Henderson and Quandt's <em>Microeconomic Theory</em> (1956).</p>

<p>My confusion lies with obtaining the total differential of the generalized utility function after substituting in the budget constraint. For example, suppose you have the following utility function:</p>

<p>$$U=f(q_1,q_2)$$
Subject to:
$$y^0=p_1q_1+p_2q_2$$</p>

<p>The total differential of this utility function is:</p>

<p>$$dU=f_{q_1}dq_1+f_{q_2}dq_2$$</p>

<p>where $f_{q_1}$ and $f_{q_2}$ are the partial derivatives of $U$ with respect to $q_1$ and $q_2$, and $dq_1$ and $dq_2$ are the changes in $q_1$ and $q_2$.  

<p>Substituting the budget constraint into the utility function results in:
$$U=f(q_1,\frac{y^0-p_1q_1}{p_2})$$</p>

<p>I am struggling to calculate the total differential of the above function. Would the total differential just be the partial derivative of $U$ with respect to $q_1$ multiplied by the change in $q_1$? Below is my attempt:</p>

<p>$$U=f(q_1,\frac{y^0}{p_2}-\frac{p_1}{p_2}q_1)$$
$$dU=f_{q_1}dq_1-\frac{p_1}{p_2}dq_1$$</p>

<p>The next step in the book (after skipping the above total differential) is:
$$\frac{dU}{dq_1}=f_1+f_2(-\frac{p_1}{p_2})=0$$</p>

<p>Obviously, my intermediate calculation is incorrect (probably in multiple ways). Can someone identify and clarify my mistakes in progressing from the substituted utility function to the above equation? I understand that this process is used to equate the marginal rate of substitution to the goods price ratio and I am aware of its interpretation. My question lies simply with the mathematical derivation.</p>

<microeconomics><mathematical-economics>
Total Differential of a Utility Function Subject to a Budget Constraint
Amaziah
# Total Differential of a Utility Function Subject to a Budget Constraint

(Asked by: username_0 on 2017-03-18)

I am trying to fully understand the process of maximizing a utility function subject to a budget constraint while utilizing the Substitution Method (as opposed to the Lagrangian Method). I am following the work of Henderson and Quandt's *Microeconomic Theory* (1956).

My confusion lies with obtaining the total differential of the generalized utility function after substituting in the budget constraint. For example, suppose you have the following utility function:

$$U=f(q\_1,q\_2)$$
Subject to:
$$y^0=p\_1q\_1+p\_2q\_2$$

The total differential of this utility function is:

$$dU=f\_{q\_1}dq\_1+f\_{q\_2}dq\_2$$

where $f\_{q\_1}$ and $f\_{q\_2}$ are the partial derivatives of $U$ with respect to $q\_1$ and $q\_2$, and $dq\_1$ and $dq\_2$ are the changes in $q\_1$ and $q\_2$.

Substituting the budget constraint into the utility function results in:
$$U=f(q\_1,\frac{y^0-p\_1q\_1}{p\_2})$$

I am struggling to calculate the total differential of the above function. Would the total differential just be the partial derivative of $U$ with respect to $q\_1$ multiplied by the change in $q\_1$? Below is my attempt:

$$U=f(q\_1,\frac{y^0}{p\_2}-\frac{p\_1}{p\_2}q\_1)$$
$$dU=f\_{q\_1}dq\_1-\frac{p\_1}{p\_2}dq\_1$$

The next step in the book (after skipping the above total differential) is:
$$\frac{dU}{dq\_1}=f\_1+f\_2(-\frac{p\_1}{p\_2})=0$$

Obviously, my intermediate calculation is incorrect (probably in multiple ways). Can someone identify and clarify my mistakes in progressing from the substituted utility function to the above equation? I understand that this process is used to equate the marginal rate of substitution to the goods price ratio and I am aware of its interpretation. My question lies simply with the mathematical derivation.

**Tags:** microeconomics, mathematical-economics

---

## 1 answers

### Answer by username_1 on 2017-03-19. Score: 2 | Accepted answer

In the second term of the intermediate calculation we must find the partial derivative of the entire second parameter of the $f$ function. Hence
$$dU = \frac{\partial f}{\partial{q\_1}}dq\_1+\frac{\partial f}{\partial (\frac{y\_0-p\_1q\_1}{p\_2})}d(\frac{y\_0-p\_1q\_1}{p\_2})$$. Obviously this is a substitution of $q\_2$ into the original total differential derivation you had. My point is that you tried to do the partial differential of $f$ with respect $q\_2$ by assuming that this would equal $$\frac{dq\_2}{dq\_1}$$ when this may not always be the case. For the final equation that you have may want to consider the chain rule. The way you have it, we have defined $$U=f(q\_1, q\_2)$$ and $$q\_2(q\_1)=\frac{y\_0}{p\_2}-\frac{p\_1}{p\_2}$$ Then according to the chain rule you have $$\frac{dU}{dq\_1}=\frac{\partial f}{\partial q\_1}\frac{dq\_1}{dq\_1}+\frac{\partial f}{\partial q\_2}\frac{dq\_2}{dq\_1}$$ Then we evaluate $$\frac{dq\_1}{dq\_1}=\frac{d}{q\_1}q\_1=1$$ Then we get $$\frac{dU}{dq\_1}=\frac{\partial f}{\partial q\_1}+\frac{\partial f}{\partial q\_2}\frac{dq\_2}{dq\_1}$$ Then, $$\frac{dq\_2}{dq\_1}=-\frac{p\_1}{p\_2}$$ Finally we end up with $$\frac{dU}{dq\_1}=f\_1+f\_2(-\frac{p\_1}{p\_2})$$


---

