How to prove that if we have $f(x)=g(y)$, then we have $f'(x)=g'(y)$ In other words, how can we apply differential operator $d$ to both side of the equation when there are different variables on both side? Thanks!

Update: Sorry for the confusion. I saw in a tutorial that:
$$\ln y = a + b\ln x \Rightarrow \frac{dy}{y} = b\frac{dx}{x}$$
and I was confused by this step. Seems to me that if $f(x)=g(y)$, then we have $f'(x)=g'(y)$. 
The link to the tutorial is here. The calculation is at around $2'20"$. Here is the screen capture:

Thanks a lot.
 A: Physicists (and the alike) like to threat the derivative $\frac{dy}{dx}$ as a fraction and multiply the relation 
$$\frac{\frac{dy}{dx}}{y} = b\frac{1}{x}$$
by '$dx$' to get
$$\frac{dy}{y} = b\frac{dx}{x}$$
This is what I think you do when you say 'apply the differential operator $d$'. The interpretation of this is simply 'if we make an very small change $dx$ in $x$ the corresponding change $dy$ in $y$ is $\ldots$'. The mathematical justification for why we can do this is that the change in $y$, $\Delta y$, if we make a small change $\Delta x$ in $x$ is given by
$$\Delta y = y(x+ \Delta x) - y(x) \approx \frac{dy}{dx} \Delta x $$
whenever $|\Delta x| \ll 1$. Thus $\frac{\Delta y}{\Delta x} \approx \frac{dy}{dx}$. The left hand side of this last equation is a real fraction, but the derivative $\frac{dy}{dx}$ is a not a fraction. However because of the notation for the derivative it kind of acts like it is. This procedure of treating $\frac{dy}{dx}$ as a fraction is not rigorous mathamtics, but it makes intuitive sense, is very helpful in solving physics problems fast and it almost always gives the correct result (if you know what you are doing).
Now to answer your question. If we have an equation with some functions that depends on several variables which all depends on the same underlying variable (which is usually time) then you can do this procedure on both sides of the equation. Lets say we have 
$$\log y(x_1,x_2) = a + b_1 \log(x_1) + b_2 \log(x_2)$$
where $x_1 = x_1(t)$, $x_2 = x_2(t)$. Lets first start with the mathematically rigorous derivation of this. We start by taking the total derivative with respect to that underlying variable $t$:
$$\frac{d\log y}{dt} = \frac{dx_1}{dt}\frac{\partial}{\partial x_1}(a + b_1 \log(x_1) + b_2 \log(x_2)) + \frac{dx_2}{dt}\frac{\partial}{\partial x_2}(a + b_1 \log(x_1) + b_2 \log(x_2))$$
to get
$$\frac{\frac{dy}{dt}}{y} = \frac{dx_1}{dt}\frac{b_1}{x_1} + \frac{dx_2}{dt}\frac{b_2}{x_2}$$
We now threat derivatives as fractions (in the sense and with the interpretation as explained above) to get
$$\frac{dy}{y} = dx_1\frac{b_1}{x_1} + dx_2\frac{b_2}{x_2}$$
This is the same as you get by the physics-procedure you do when you say that you 'apply the differential operator $d$'. Note that it is very important here that we only have only one fundamental variable (here it's $t$) that all the other variables depend on. If we had more then the standard physics-procedure as outlined here would not work (well, strickly speaking it can work - and one does this in thermodynamics - but it's more complicated that outlined here).
To your second question: If we have $f(y) = g(x)$ then is $f'(y)=g'(x)$? To answer this (and for this to even make sense) you first have to specify what the derivative is with respect to and what the variables depend on. I will assume that you mean that $y$ depend on $x$ where $x$ is the fundamental variable and that the derivative is with respect to $x$. Then yes, if $f(y(x))=g(x)$ then $\frac{df(y(x))}{dx} = \frac{dg(x)}{dx}$.
