# Under what conditions can a function $y: \mathbb{R} \to \mathbb{R}$ be expressed as $\dfrac{z'}{z}$?

Can an arbitrary function $y: \mathbb{R} \to \mathbb{R}$ always be expressed as $\dfrac{z'}{z}$ for some differentiable function $z: \mathbb{R} \to \mathbb{R}$, or are additional conditions on $y$ needed for this to be true? This question was inspired by reading a writeup about differential equations, where the change of variables $y = \dfrac{z'}{z}$ is made.

• Think about $z(x) = e^{\int y(x) ~ \mathrm{d}{x}}$. Jun 15, 2014 at 18:28

## 1 Answer

Let $y: \mathbb{R} \to \mathbb{R}$ be any continuous function. Define $z: \mathbb{R} \to \mathbb{R}$ by $$\forall x \in \mathbb{R}: \quad z(x) \stackrel{\text{def}}{=} e^{F(x)},$$ where $F$ is an antiderivative of $y$ (which exists). Notice that $z$ is a positive function. By the Exponential Chain Rule, $$\forall x \in \mathbb{R}: \quad z'(x) = e^{F(x)} \cdot y(x).$$ Therefore, $$\forall x \in \mathbb{R}: \quad y(x) = \dfrac{e^{F(x)} \cdot y(x)}{e^{F(x)}} = \dfrac{z'(x)}{z(x)}.$$

Additional Set-Theoretic Information

Let $y: \mathbb{R} \to \mathbb{R}$, and suppose that there exists a differentiable $z: \mathbb{R} \to \mathbb{R}^{+}$ such that $y = \dfrac{z'}{z}$. As $\dfrac{z'}{z} = (\ln \circ z)'$, it follows that $y$ is a derivative (or equivalently, $y$ has an anti-derivative). Then according to the rather fantastic answer in this post, the set of discontinuities of $y$ is a meager $F_{\sigma}$-subset of $\mathbb{R}$.

Conversely, if the set of discontinuities of $y$ is a meager $F_{\sigma}$-subset of $\mathbb{R}$, then $y$ has an anti-derivative $F$, which can be used to construct the required function $z$ in the manner above.

Conclusion: Let $y: \mathbb{R} \to \mathbb{R}$. Then there exists a differentiable $z: \mathbb{R} \to \mathbb{R}^{+}$ such that $y = \dfrac{z'}{z}$ if and only if the set of discontinuities of $y$ is a meager $F_{\sigma}$-subset of $\mathbb{R}$.

• Interesting...this means then than a function y(x) can be expressed as z'(x)/z(x) for some z iff it can be expressed as z'(x) for some z. Wonder if that is the case when one replaces z'(x)/z(x) by z'(x)*z(x), just asked it as a different question: math.stackexchange.com/questions/835774/… Jun 16, 2014 at 7:12