6
$\begingroup$

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?

$\endgroup$
12
  • 12
    $\begingroup$ The answer is just 1! $\endgroup$
    – danimal
    May 8, 2015 at 10:02
  • 7
    $\begingroup$ No, definitely 1 $\endgroup$
    – danimal
    May 8, 2015 at 10:04
  • 8
    $\begingroup$ @GitGud No, it's 1 $\endgroup$ May 8, 2015 at 10:48
  • 5
    $\begingroup$ @AlexandreHalm The same theory in which $dy/dx$ or $dx/dx$ do. $\endgroup$ May 8, 2015 at 11:53
  • 5
    $\begingroup$ 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, 2015 at 6:22

5 Answers 5

20
$\begingroup$

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, $\displaystyle\frac{d}{dx}(y)=\frac{d}{dx}(x)=1$.

$\endgroup$
4
  • 1
    $\begingroup$ 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. $\endgroup$ May 8, 2015 at 17:26
  • 6
    $\begingroup$ If $y$ is meant to be a variable, how can it be constant? It would be of no use in differentiation. $\endgroup$
    – yellon
    May 8, 2015 at 18:54
  • 3
    $\begingroup$ 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, 2015 at 21:56
  • 2
    $\begingroup$ @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. $\endgroup$ May 9, 2015 at 8:10
18
$\begingroup$

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.

$\endgroup$
3
  • 8
    $\begingroup$ Plus 1 for first principles $\endgroup$
    – danimal
    May 8, 2015 at 15:28
  • $\begingroup$ What if y is a constant and so Δy equals zero? $\endgroup$ May 8, 2015 at 17:28
  • 3
    $\begingroup$ 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, 2015 at 21:35
2
$\begingroup$

$$\frac d{dy}$$ 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 $y$.)

$\endgroup$
3
  • $\begingroup$ What if y is a constant and so Δy equals zero? $\endgroup$ May 8, 2015 at 17:29
  • 1
    $\begingroup$ @StevenStewart-Gallus You could say the same thing to $\frac{dy}{dx}$. What if $x$ is constant? $\endgroup$ May 8, 2015 at 17:35
  • 1
    $\begingroup$ I agree with colombus: Why would one ever differentiate anything, constant or otherwise, with respect to a constant? Objection overruled. $\endgroup$
    – pjs36
    May 8, 2015 at 17:58
2
$\begingroup$

Yes , example y=y, dy/dy=1 , means that y is equal to its own coordinate and its rate of change is 1

$\endgroup$
9
  • 1
    $\begingroup$ what does this mean in the context of the problem?? $\endgroup$
    – Bak1139
    May 8, 2015 at 10:05
  • $\begingroup$ Ignore it misread question $\endgroup$
    – talmudist
    May 8, 2015 at 10:07
  • 4
    $\begingroup$ @talmudist You can delete answers, אחי. $\endgroup$ May 8, 2015 at 11:18
  • $\begingroup$ How can I delete answer ? No option for it $\endgroup$
    – talmudist
    May 8, 2015 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, 2015 at 13:18
0
$\begingroup$

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.

$\endgroup$
1
  • 4
    $\begingroup$ 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. $\endgroup$ May 8, 2015 at 19:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.