Why $\frac{y’}{y}$ is equivalent to $(\ln|y|)’$?

Question

I just read a proof process from a textbook about differential equations, it has following content : $$y’= ay\implies\frac{y’}{y}= a\implies (\ln|y|)’= a$$ I want to know why $\frac{y’}{y}$ is equivalent to $(\ln|y|)’$?

Answer 1

$\frac{d}{dx}[\ln|y|]=\frac{d}{dx}[\ln|x|]\bigg|_y\cdot \frac{d}{dx}[y]=\frac{1}{y}\cdot \frac{d}{dx}[y] = \frac{y’}{y}$

By the Chain Rule, because $\frac{d}{dx}[\ln|x|]=\frac{1}{x}$.

As for the absolute value, this link will provide context, $\ln(y)$ is defined for $y>0$ only, the absolute value allows you to consider negative values of $y\neq 0$.

Answer 2

Let $f$ be the function defined by $f(y)=\log|y|$ for all $y\neq0$. We see that $$ f(y)= \begin{cases} \log(y) &\text{if $y>0$} \\ \log(-y) &\text{if $y<0$} \, . \end{cases} $$ If $y>0$, then $$\frac{d}{dy}f(y)=\frac{d}{dy}(\log y)=\frac{1}{y} \, .$$ And if $y<0$, then $$\frac{d}{dy}f(y)=\frac{d}{dy}(\log(-y))=\frac{1}{-y} \cdot -1=\frac{1}{y}$$by the chain rule. Hence, $f'(y)=1/y$ for all $y\neq0$. By contrast, the function $g$ defined by $g(y)=\log y$ has the domain $\mathbb{R}^+$ and so $g'(y)=1/y$ only when $y>0$. Hence, $g$ is not an antiderivative of the function $h$ given by $h(y)=1/y$, since the domain of $h$ is $\mathbb{R} \setminus \{0\}$. So while we have $f'=h$, we don't have $g'=h$.

Answer 3

$$y'=ay$$ $$\dfrac {dy}{dx}=ay$$ This is separable: $$\int \dfrac {dy}{y}=a\int dx$$ $$\ln y=ax+C$$ Differentiate: $$(\ln y)'=a$$