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: Yes , example y=y, dy/dy=1 , means that y is equal to its own coordinate and its rate of change is 1  
A: $$\frac d{dy}$$
means differentiation with respect to $y$. Thus,
$$\frac{dy}{dy}$$
is differentiating $y$ with respect to $y$. The derivative of $y$ with respect to $y$ is $\boxed1$. (It's slightly easier to understand if you replace all of the $y$s with $x$ — we're more used to differentiating with respect to $x$ than with respect to $y$.)
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.
A: dy/dy is 1 unless possibly y is a constant or has constant portions. However, this statement seems kind of meaningless. What is y a constant in respect to?
One possible way to solve the problem might be to formulate everything as varying according to a single global time parameter t. Then dy/dy is better described as dyₜ/dyₜ and by the chain rule is dyₜ/dt * dt/dy. dyₜ/dt is simply 0 when y is constant with respect to 0 but finding a useful definition for dt/dyₜ is harder. As yₜ does not vary dt/dyₜ is undefined or approaches a positive infinity. As such, dyₜ/dyₜ is best described as being undefined, I think. Alternatively it is not appropriate to use the chain rule when the denominator is a constant  value. However, some people may find it useful to patch in an ugly hack to our normal definition of the derivative and arbitrarily define dyₜ/dyₜ to be 1 or some other useful value for their purposes. I am not necessarily convinced such a patch would be useful but I would be open to the possibility.
