# Where does this step into solving an equation come from?

I know this might sound dumb, but becuause I am a "beginner" i have this doubt a long time ago, and I would like a complete exlpanation about why it is possible to do it. And most because I have seen that too many physicists use it.

It is about the following step involving solving this problem: $$\dfrac{df(x)}{dx}=\dfrac{1}{x}$$ $$\Rightarrow df(x)=\dfrac{dx}{x}$$ $$\Rightarrow f(x)=ln(x)+k$$ being k a constant. It is just because of the first and second fundamental calculus theorem? But what happen if I work with several variables?

Starting from: $$\dfrac{df(x)}{dx}=\dfrac{1}{x}$$ you could say you then integrate both sides with respect to $x$, that is: $$\color{blue}{\int} \dfrac{df(x)}{dx} \,\color{blue}{dx}=\color{blue}{\int}\dfrac{1}{x}\,\color{blue}{dx}$$ Now the left-hand side is, due to the fundamental theorem of calculus: $$\int \dfrac{df(x)}{dx} \, dx = \int f'(x) \, dx = f(x) \quad(+C)$$ You can integrate the right-hand side with the usual rules for integration.
In practice, more particularly in the context of differential equations, the following notation is often used when applying separation of variables. Split the $y=f(x)$ and the $x$ parts as you wrote it and follow up with "integrating both sides": $$\dfrac{df(x)}{dx}=\dfrac{1}{x} \quad\rightarrow\quad df(x)=\dfrac{1}{x} dx \quad\rightarrow\quad \color{blue}{\int}df(x)=\color{blue}{\int}\dfrac{1}{x} dx$$
• A function defined as $f:\Bbb R^n \rightarrow \Bbb R^m$, i mean how far does this "trick" work, when working with several variables I would need to consider several integrations? – Mounice Dec 8 '16 at 20:13