Q:

Suppose $y$, Does $\frac{dy}{dy} $ have meaning when we derive it with respect to itself?

Suppose we have a function and want to derive it with respect to itself e.g: $$\frac{dy}{dy} $$
Does this have any meaning , and if so what will be it's value? 

A:

We are basically asking what is the rate by which $y$ changes with respect to $y$? Since $y$ changes proportionately to itself, the value is $1$.
Notice that you probably do this implicitly if you have differentiated some functions before. For instance, if $y=2x$, then $\displaystyle\frac{d}{dx}(y)=\frac{d}{dx}(2x)=2.$ That is, the rate by which $y$ changes with respect to $x$ is $2$, since every time $y$ increases or decreases by $2$, $x$ only increases or decreases by $1$. Now suppose instead that $y=x$, since $y$ is changing only as much as $x$ is, the rate by which $y$ changes with respect to $x$ is now $1$. Thus, $\displaystyle\frac{d}{dx}(y)=\frac{d}{dx}(x)=1$.

A:

Define the function $f(y)=y$ $\;\;\;\;\;\;$ (1)
We will use the first principle here, 
Let a small increment in $y$ correspond to $f(y+\Delta y)$
Then , $f(y+\Delta y)=y+\Delta y$ $\;\;\;\;\;\;\;$ (2)
Subtract (1) from (2)
$f(y+\Delta y)-f(y)=\Delta y$
Re-arranging. 
$\frac{f(y+\Delta y)-f(y)}{\Delta y}=1$
Taking limit $\Delta y\to 0$
$\frac{d[f(y)]}{dy}=1=\frac{d(y)}{dy}$
Hence, essentially you are measuring the rate of change of quantity with respect to itself. It's quite easy to see why it makes sense.

