Working with the $\frac{d}{dx}$ operator I have a fundamental query about the way derivatives can be used in algebraic manipulations. 
Say $\dfrac{d(\ln x)}{dx}=\dfrac{1}{x}$
Apparently, this can be manipulated to $d(\ln x)=\dfrac{dx}{x}$. I understand that this can be integrated back to the first equation.
But, the reason this was done was to depict $\dfrac{dx}{x}$ as a percentage change in $x$.
The definition of $\dfrac{df(x)}{dx}=\lim_\limits{h\to 0}\dfrac{f(x+h)-f(x)}{h}$. So, isn't $\dfrac{d}{dx}$ more like a function than a ratio of two quantities ?
If so, how is moving $dx$ to RHS possible as done above ?
Please advise.
 A: Mathematically speaking, $\frac{d}{dx}$ is an operator whose work is to differentiate functions, just like $+,-,*,/$ are all operators whose respective functions are to add, subtract, multiply and divide.
The multiplication of $dx$ however proceeds as follows:
$$\frac{d}{dx}f(x)=f'(x) \Rightarrow \frac{d}{dx}f(x) dx=f'(x) dx \Rightarrow d\{f(x)\}=f'(x) dx$$
Now it is logically deduced with the help of infinitesimals that $$\frac{d}{dx}f(x) dx=d\{f(x)\}$$
It is not that the $dx$ in the numerator and denominator just cancel out like a ratio. 
A: Intuitively, $dx$ is an infinitely small change in the value of $x$.  This results in an infinitely small change in $\log x$, and that change is $d(\log x) = \dfrac{dx} x$.
If we then go on to conclude that
$$
\int_{[a,b]} g \cdot \log' =\int_a^b g(x) \log' x\,dx = \int_a^b \frac{g(x)} x\,dx
$$
then the identity is correct, and essentially that's an application of the chain rule.
A: I have gone through the same problem when I was learning calculus for the first time. Let me explain how I understood the differential sign $\frac{d}{dx}$. basically the basic concept I understood was;

derivatives are nothing but SLOPES of a certain function.

I started with the definition of slopes which is $\frac{\Delta y}{\Delta x }$, assuming my $\Delta$ simply to mean "significant change or big change in some value".
So, the slope of $f(x)$ at some $x$ is simply,
$$\frac{\Delta f(x)}{\Delta x} .$$ I asked myself: "can I consider this as a ratio and simply manipulate them as a ratio ($\frac{a}{b}=1\Rightarrow a=b$). I came to the conclusion that I could do so by taking the example of $f(x)$ being a straight line (we use this method if two points are given and we need to find the equation of $y$).
Now that I did this I came back to $\frac{d}{dx}$. So the definition states that $$\frac{df(x)}{dx}=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}.$$  So this came from the fact that $h=x_{1}-x$, where $x_{1}$ is a variable and $x$ is some constant we want to find the slope at. Replacing we get,
$$\lim_{x_{1}\rightarrow x}\frac{f(x_{1})-f(x)}{x_1-x}.$$
This is the first slope I discussed, it's just that $x_{1}$ and $x$ are so close to each other, so $\frac{df(x)}{dx}$ simply meant slope with $d$ denoting small change or insignificant change. Since I understood that it is nothing but slope which by definition is ratio I understood I could manipulate them.
A: I find it most beneficial to view $\frac{d}{dx}$ as being two operations - a differential and a division.  Most of Calculus notation becomes much more clear when you think of it that way.  Then the terms $dx$ and $dy$ can be manipulated as algebraic terms.
