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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?

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    $\begingroup$ The answer is just 1! $\endgroup$ – danimal May 8 '15 at 10:02
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    $\begingroup$ No, definitely 1 $\endgroup$ – danimal May 8 '15 at 10:04
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    $\begingroup$ @GitGud No, it's 1 $\endgroup$ – Akiva Weinberger May 8 '15 at 10:48
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    $\begingroup$ @AlexandreHalm The same theory in which $dy/dx$ or $dx/dx$ do. $\endgroup$ – Akiva Weinberger May 8 '15 at 11:53
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    $\begingroup$ Just curious, but any particular reason for the rollback of my edit? (However, if you do not wish to explain, please feel free to ignore this comment.) $\endgroup$ – J W May 9 '15 at 6:22
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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, $\displaystyle\frac{d}{dx}(y)=\frac{d}{dx}(x)=1$.

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    $\begingroup$ Possibly wrong if dy is zero (y is a constant). I am not sure what the most useful and appropriate definition to give dy/dy when y is constant is though. $\endgroup$ – Steven Stewart-Gallus May 8 '15 at 17:26
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    $\begingroup$ If $y$ is meant to be a variable, how can it be constant? It would be of no use in differentiation. $\endgroup$ – yellowquark May 8 '15 at 18:54
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    $\begingroup$ Basically writing $\frac{dx}{dx}$ is an abuse of notation. When writing $\frac{dx}{dx}$ what you actually mean is $\frac{df}{dx}$ when $f=\text{identity}$, i.e., $f(x)=x$. $\endgroup$ – hjhjhj57 May 8 '15 at 21:56
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    $\begingroup$ @hjhjhj57 If you want to take this stance, then even writing $\frac{dy}{dx}$ is an abuse of notation, because it is nowhere indicated that $y$ is a function of $x$. The most "correct" way would be if you indicate that $D$ is an operator on functions, i.e. something like $D[(x\in\Bbb R\mapsto x)]=(x\in\Bbb R\mapsto 1)$, but needless to say this notation isn't likely to catch on anytime soon. $\endgroup$ – Mario Carneiro May 9 '15 at 8:10
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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.

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    $\begingroup$ Plus 1 for first principles $\endgroup$ – danimal May 8 '15 at 15:28
  • $\begingroup$ What if y is a constant and so Δy equals zero? $\endgroup$ – Steven Stewart-Gallus May 8 '15 at 17:28
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    $\begingroup$ If $y$ is constant, then you have a difficulty defining $d(\mbox{anything})/dy$. Every derivative has this problem; it is nothing special about $dy/dy$. $\endgroup$ – David K May 8 '15 at 21:35
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$$\frac d{dy}$$ 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 $y$.)

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  • $\begingroup$ What if y is a constant and so Δy equals zero? $\endgroup$ – Steven Stewart-Gallus May 8 '15 at 17:29
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    $\begingroup$ @StevenStewart-Gallus You could say the same thing to $\frac{dy}{dx}$. What if $x$ is constant? $\endgroup$ – Akiva Weinberger May 8 '15 at 17:35
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    $\begingroup$ I agree with colombus: Why would one ever differentiate anything, constant or otherwise, with respect to a constant? Objection overruled. $\endgroup$ – pjs36 May 8 '15 at 17:58
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Yes , example y=y, dy/dy=1 , means that y is equal to its own coordinate and its rate of change is 1

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    $\begingroup$ what does this mean in the context of the problem?? $\endgroup$ – Bak1139 May 8 '15 at 10:05
  • $\begingroup$ Ignore it misread question $\endgroup$ – talmudist May 8 '15 at 10:07
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    $\begingroup$ @talmudist You can delete answers, אחי. $\endgroup$ – Akiva Weinberger May 8 '15 at 11:18
  • $\begingroup$ How can I delete answer ? No option for it $\endgroup$ – talmudist May 8 '15 at 11:57
  • $\begingroup$ @talmudist underneath the text of your post, you should see text links that say 'share' 'edit' 'delete' 'flag' $\endgroup$ – AakashM May 8 '15 at 13:18
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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.

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    $\begingroup$ Possibly wrong if the global time parameter t that you used is constant. Also what does "when y is constant with respect to 0" mean? I think the piece you are missing is that Leibnitz notation implicitly uses the quantity that appears in the denominator as a variable. So the worry about what happens when y is constant is a red herring. As others have pointed out you would have the same problem with dy/dx worrying about what happens when x is constant as well. $\endgroup$ – BSteinhurst May 8 '15 at 19:18

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