<p>As Herr K. pointed out, if you write <span class="math-container">$\frac{du(g)}{dg}$</span>, then <span class="math-container">$u$</span>'s domain has to be <span class="math-container">$\mathbb{R}^+$</span> and <span class="math-container">$x,z$</span> have to be real numbers.</p>
<p>You'd rather ask:</p>
<blockquote>
<p>When <span class="math-container">$u : \mathbb{R}^+ \to \mathbb{R}^+$</span> increases along with <span class="math-container">$d(x,z) := |x-z|$</span>, can we say <span class="math-container">$\frac{du(g)}{dg} &gt; 0$</span>?</p>
</blockquote>
<p>Yes, if <span class="math-container">$u(x)$</span> is differentiable at all points in its domain.</p>
<p>Since <span class="math-container">$\text{range}(\{d(x,z) : x,z \in \mathbb{R}^+\}) = \mathbb{R}^+$</span>, <span class="math-container">$\frac{du(g)}{dg}$</span> is equivalent to writing <span class="math-container">$u'(x)$</span> where <span class="math-container">$x \in \mathbb{R}^+$</span>. For you to be able to define this way, <span class="math-container">$u$</span> has to be differentiable throughout its domain which may not always be the case.</p>
<p>A better way to verify if <span class="math-container">$u$</span> increases with <span class="math-container">$d$</span> would be to check if <span class="math-container">$u$</span> is a (strictly) increasing function which doesn't require differentiability.</p>

CC BY-SA 4.0
2022-06-24T05:39:34.773Z
Can you explain why $u$'s domain has to be $\mathbb{R}^+$? Does this not imply we are in a one-dimensional space? I would like to make the statement in respect to two vectors. i.e the Euclidean distance between 2 points in a two-dimensional space, $\mathbb{R}_+^2$. The distance would then be found using the square-root formula.
Eli J
CC BY-SA 4.0
2022-06-24T06:52:28.917Z
@EliJ For $x,z \in \mathbb{R}^+$, $g(x,z) = |x-z| \in \mathbb{R^+}$. When you have $g$ inside $u$, then the range of $g$ forms the domain of $u$. And the range is $\mathbb{R}^+$ as I wrote in the post.
Kur_Kush
CC BY-SA 4.0
2022-06-24T07:01:29.890Z
@EliJ As for using "two vectors" where each vector lies in the 2D plane, you can use the function $v(x,z) := (u \circ g)(x,z)$ where $v : \mathbb{R_+}^2 \times \mathbb{R_+}^2 \to \mathbb{R_+}$.
Kur_Kush
CC BY-SA 4.0
2022-06-24T04:10:55.757Z
2022-06-24T04:10:55.757Z
Kur_Kush
<p>I have a utility function <span class="math-container">$u(x,z)$</span> from <span class="math-container">$\mathbb{R}_+$</span> to <span class="math-container">$\mathbb{R_+}$</span>,  where <span class="math-container">$x,z \in \mathbb{R}_+$</span>.</p>
<p>I would like to turn the following statement into math: &quot;the utility function <span class="math-container">$u$</span> is increasing as the Euclidean distance between <span class="math-container">$x$</span> and <span class="math-container">$z$</span> is increasing&quot;.</p>
<p>Can I write <span class="math-container">$\frac{ d u(g) }{d g}&gt;0$</span>, where <span class="math-container">$g=d(x,z)\in \mathbb{R_+} $</span> is the Euclidean distance between <span class="math-container">$x$</span> and <span class="math-container">$z$</span>?</p>

CC BY-SA 4.0
2022-06-24T03:26:20.310Z
I think you mean $x,z\in\mathbb R_+$ in the first paragraph.
Herr K.
CC BY-SA 4.0
2022-06-24T03:26:39.293Z
$\frac{\mathrm du(g)}{\mathrm dg}>0$ makes sense, but if $g$ is the only argument in $u$, then you should probably change $u$'s domain from $\mathbb R_+^2$ to $\mathbb R_+$. Perhaps define another function $v(x,z)=u(g(x,z))$.
Herr K.
<mathematical-economics>
A derivative with respect to Euclidean distance?
Eli J
# A derivative with respect to Euclidean distance?

(Asked by: username_0 on 2022-06-23)

I have a utility function $u(x,z)$ from $\mathbb{R}\_+$ to $\mathbb{R\_+}$, where $x,z \in \mathbb{R}\_+$.

I would like to turn the following statement into math: "the utility function $u$ is increasing as the Euclidean distance between $x$ and $z$ is increasing".

Can I write $\frac{ d u(g) }{d g}>0$, where $g=d(x,z)\in \mathbb{R\_+} $ is the Euclidean distance between $x$ and $z$?

**Tags:** mathematical-economics

> **Comments:**
> **username_1**: I think you mean $x,z\in\mathbb R_+$ in the first paragraph.  
> **username_1**: $\frac{\mathrm du(g)}{\mathrm dg}>0$ makes sense, but if $g$ is the only argument in $u$, then you should probably change $u$'s domain from $\mathbb R_+^2$ to $\mathbb R_+$. Perhaps define another function $v(x,z)=u(g(x,z))$.  

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## 1 answers

### Answer by username_2 on 2022-06-24. Score: 2 | Accepted answer

As username_1. pointed out, if you write $\frac{du(g)}{dg}$, then $u$'s domain has to be $\mathbb{R}^+$ and $x,z$ have to be real numbers.

You'd rather ask:

> When $u : \mathbb{R}^+ \to \mathbb{R}^+$ increases along with $d(x,z) := |x-z|$, can we say $\frac{du(g)}{dg} > 0$?

Yes, if $u(x)$ is differentiable at all points in its domain.

Since $\text{range}(\{d(x,z) : x,z \in \mathbb{R}^+\}) = \mathbb{R}^+$, $\frac{du(g)}{dg}$ is equivalent to writing $u'(x)$ where $x \in \mathbb{R}^+$. For you to be able to define this way, $u$ has to be differentiable throughout its domain which may not always be the case.

A better way to verify if $u$ increases with $d$ would be to check if $u$ is a (strictly) increasing function which doesn't require differentiability.

> **Comments:**
> **username_0**: Can you explain why $u$'s domain has to be $\mathbb{R}^+$? Does this not imply we are in a one-dimensional space? I would like to make the statement in respect to two vectors. i.e the Euclidean distance between 2 points in a two-dimensional space, $\mathbb{R}_+^2$. The distance would then be found using the square-root formula.  
> **username_2**: @username_0 For $x,z \in \mathbb{R}^+$, $g(x,z) = |x-z| \in \mathbb{R^+}$. When you have $g$ inside $u$, then the range of $g$ forms the domain of $u$. And the range is $\mathbb{R}^+$ as I wrote in the post.  
> **username_2**: @username_0 As for using "two vectors" where each vector lies in the 2D plane, you can use the function $v(x,z) := (u \circ g)(x,z)$ where $v : \mathbb{R_+}^2 \times \mathbb{R_+}^2 \to \mathbb{R_+}$.  

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