# A derivative with respect to Euclidean distance?

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 $$z$$?

• I think you mean $x,z\in\mathbb R_+$ in the first paragraph. yesterday
• $\frac{\mathrm du(g)}{\mathrm dg}>0$ 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 $v(x,z)=u(g(x,z))$. yesterday

As Herr K. 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.

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.
• 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. yesterday
• @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. yesterday
• @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 $v : \mathbb{R_+}^2 \times \mathbb{R_+}^2 \to \mathbb{R_+}$. yesterday