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I recently came accross a differential equation where there was $\frac{y'}{y}$. How can we prove that $\frac{y'}{y}=(\ln y)' $?

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closed as off-topic by Did, C. Falcon, projectilemotion, ΘΣΦGenSan, Leucippus Mar 14 '17 at 1:14

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  • $\begingroup$ Just apply the chain rule. $\endgroup$ – MatheinBoulomenos Mar 11 '17 at 16:04
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Assuming $y$ is a function of another variable and can be written, $y(t)$, we have \begin{align*} \frac{d(\log y(t))}{dt} = \frac{1}{y(t)} \cdot y'(t) \end{align*} by the chain rule.

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using the chain rule we have $$(\ln(y))'=\frac{1}{y}\cdot y'$$

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