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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.

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

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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} $.

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  • $\begingroup$ 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/… $\endgroup$ Jun 16, 2014 at 7:12

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