<ol>
<li><p>I think the <span class="math-container">$\partial u$</span> is a typo as it should be:
<span class="math-container">$$
du = \frac{\partial u}{\partial z}dz + \frac{\partial u}{\partial y} dy
$$</span>
For the intuittion, consider a small change from <span class="math-container">$(z,y)$</span> to <span class="math-container">$(z + dz, y + dy)$</span>.
The change in <span class="math-container">$u$</span> is from <span class="math-container">$u(z,y)$</span> to <span class="math-container">$u(z + dz, y + dy)$</span>.
Then
<span class="math-container">$$
\begin{align*}
\underbrace{u(z + dz, y + dy)- u(z,y)}_{\approx du} &amp;= \underbrace{\frac{u(z + dz, y + dy) - u(z, y + dy)}{dz}}_{\approx\frac{\partial u}{\partial z}} dz+ \underbrace{\frac{u(z, y + dy) - u(z,y)}{dy}}_{\approx \frac{\partial u}{\partial y}}{dy},\\
du &amp;= \frac{\partial u(z,y)}{\partial z} dz + \frac{\partial u(z,y)}{\partial y} dy
\end{align*}
$$</span></p>
</li>
<li><p>The <span class="math-container">$\partial u/\partial z$</span> is indeed measures the marginal change in <span class="math-container">$u$</span> when <span class="math-container">$z$</span> increases by one unit. The <span class="math-container">$dz$</span> and <span class="math-container">$dy$</span> takes into account that <span class="math-container">$z$</span> may change more or less (relatively) compared to <span class="math-container">$y$</span>. So although <span class="math-container">$dz$</span> and <span class="math-container">$dy$</span> are both (arbitrarily) small the relative change <span class="math-container">$dz/dy$</span> might be big. For more information, you might look up the concept of a directional derivative.</p>
</li>
</ol>

CC BY-SA 4.0
2024-12-23T13:20:45.180Z
Thanks for the response, it cleared my doubts. I had another question which is game theory related. You could take a look at it, you'll probably be able to answer that one too. Here it is: https://economics.stackexchange.com/questions/59678/what-is-the-difference-between-a-nash-equilibrium-and-a-dominant-strategy
BeginnerCode776
CC BY-SA 4.0
2024-12-23T11:12:35.470Z
2024-12-23T11:12:35.470Z
tdm
<p><em>I'm a little confused with partial derivatives and total derivatives and their notation.</em></p>
<p><strong>Given:</strong></p>
<p>u(z,y) = y(z-20)^2</p>
<p><strong>Change in u is then given by the following:</strong></p>
<p>∂u = ∂u/∂z dz + ∂u/∂y dy</p>
<p><em>My questions:</em></p>
<ol>
<li><p>Why is the change in u denoted as ∂u and not du? I understand that ∂u is the partial derivative of u with respect to some variable and du is the total derivative, in the case aren't we finding the total derivative from the sum of the partial derivatives?</p>
</li>
<li><p>Also, why is the dz and dy necessary? Doesn't ∂u/∂z in itself signify that u is being partially derived with respect to z? And the same with ∂u/∂y?</p>
</li>
</ol>

<utility>
What is the difference between ∂ and d notation in the change in utility formula?
BeginnerCode776
# What is the difference between ∂ and d notation in the change in utility formula?

(Asked by: username_0 on 2024-12-23)

*I'm a little confused with partial derivatives and total derivatives and their notation.*

**Given:**

u(z,y) = y(z-20)^2

**Change in u is then given by the following:**

∂u = ∂u/∂z dz + ∂u/∂y dy

*My questions:*

1. Why is the change in u denoted as ∂u and not du? I understand that ∂u is the partial derivative of u with respect to some variable and du is the total derivative, in the case aren't we finding the total derivative from the sum of the partial derivatives?
2. Also, why is the dz and dy necessary? Doesn't ∂u/∂z in itself signify that u is being partially derived with respect to z? And the same with ∂u/∂y?

**Tags:** utility


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## 1 answers

### Answer by username_1 on 2024-12-23. Score: 1

1. I think the $\partial u$ is a typo as it should be:
   $$
   du = \frac{\partial u}{\partial z}dz + \frac{\partial u}{\partial y} dy
   $$
   For the intuittion, consider a small change from $(z,y)$ to $(z + dz, y + dy)$.
   The change in $u$ is from $u(z,y)$ to $u(z + dz, y + dy)$.
   Then
   $$
   \begin{align\*}
   \underbrace{u(z + dz, y + dy)- u(z,y)}\_{\approx du} &= \underbrace{\frac{u(z + dz, y + dy) - u(z, y + dy)}{dz}}\_{\approx\frac{\partial u}{\partial z}} dz+ \underbrace{\frac{u(z, y + dy) - u(z,y)}{dy}}\_{\approx \frac{\partial u}{\partial y}}{dy},\\
   du &= \frac{\partial u(z,y)}{\partial z} dz + \frac{\partial u(z,y)}{\partial y} dy
   \end{align\*}
   $$
2. The $\partial u/\partial z$ is indeed measures the marginal change in $u$ when $z$ increases by one unit. The $dz$ and $dy$ takes into account that $z$ may change more or less (relatively) compared to $y$. So although $dz$ and $dy$ are both (arbitrarily) small the relative change $dz/dy$ might be big. For more information, you might look up the concept of a directional derivative.

> **Comments:**
> **username_0**: Thanks for the response, it cleared my doubts. I had another question which is game theory related. You could take a look at it, you'll probably be able to answer that one too. Here it is: https://economics.stackexchange.com/questions/59678/what-is-the-difference-between-a-nash-equilibrium-and-a-dominant-strategy  

---

