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Let $h(y) = y\sqrt y$.

The derivative of it is $\frac {3y} {2 \sqrt y}$ or $\frac {3 \sqrt y} 2$, when more simplified. Although both graphs of the derivative are the same, when simplified further with the denominator rationalized, it includes zero in the domain. But if left as $\frac {3y} {2\sqrt y}$, the domain excludes zero.

Which one would be the correct domain of the derivative? $(0,\infty)$ or $[0,\infty)$

EDIT: The domain for the original function h(y) is $[0,\infty)$

harold232
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    You mistagged. The tag [tag:domain-theory] says “Do not use this tag for questions about the domain of a function”. – k.stm Oct 05 '19 at 22:26
  • oh didnt see that – harold232 Oct 05 '19 at 22:27
  • @harold232 You posted a question, and deleted it seconds before I would click the post answer button (just after I had uploaded the image). If you want to see my answer it is at https://i.stack.imgur.com/PMsL9.png If you undelete your question I could post it there, up to you. – Mirko Nov 17 '19 at 13:35

2 Answers2

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The domain is $[0,\infty)$. Note that\begin{align}h'(0)&=\lim_{t\to0}\frac{h(t)-h(0)}t\\&=\lim_{t\to0}\frac{t\sqrt t}t\\&=\lim_{t\to0}\sqrt t\\&=0.\end{align}

José Carlos Santos
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  • but including zero will make the denominator zero, and division by zero makes the function undefined – harold232 Oct 05 '19 at 22:31
  • At no point of my computations there was a division by $0$. – José Carlos Santos Oct 05 '19 at 22:33
  • $\lim\limits_{t\to0^-} h(t)$ does not exist, so how can the limit exist? – Andrew Chin Oct 05 '19 at 22:33
  • I think you looked the wrong question, I was asking about the domain for the derivative not the original function.[0,∞) for the original function is correct, but im unsure wheter the domain for the derivative is [0,∞) or (0,∞) – harold232 Oct 05 '19 at 22:34
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    Asserting that $\lim_{x\to0}f(x)=l$ means that$$(\forall\varepsilon>0)(\exists\delta>0):\lvert x\rvert<\delta\implies\bigl\lvert f(x)-l\bigr\rvert<\varepsilon.$$This makes sense even if $f$ is defined only in $(0,\infty)$, as in my case. – José Carlos Santos Oct 05 '19 at 22:36
  • If you look at the derivative I typed in my question, you will see that there is a radical in its denominator, so it must be greater than 0, which excludes the typical square root domain where it CAN equal to 0. This is because division by 0 is not allowed. But when you simplify the derivative further, some reason the domain changes to including zero – harold232 Oct 05 '19 at 22:36
  • @AndrewChin : If the domain is $[0,\infty)$ then the derivative at $0$ is assumed to be the value at $0^+$. – Steven Alexis Gregory Oct 05 '19 at 22:36
  • No, I did not look at the wrong question. I am asserting that the domain of $h'$ is $[0,\infty)$. – José Carlos Santos Oct 05 '19 at 22:37
  • Anyways my only question is whether the derivative of h(y)=y√y is (0,∞) or [0,∞)? When I calculate the derivative I get (0,∞) as the domain because there is a radical in the denominator. But when I rationalize the denominator I get [0,∞) as the domain. Both derivatives are EQUIVALENT but why does the domain change when simplifying the derivative? In other words, which domain would be correct – harold232 Oct 05 '19 at 22:40
  • As I wrote in my previous comment, the domain of $h'$ is $[0,\infty)$. – José Carlos Santos Oct 05 '19 at 22:42
  • I don't understand why it is [0,∞). Both the derivatives are equivalent, but why take this domain rather than (0,∞). The original calculated derivative's domain with radical in the denominator does not include zero. Because you cannot divide by 0 – harold232 Oct 05 '19 at 22:44
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    In my answer I computed $h'(0)$. I got $0$. If there is some error in my computation, just tell me where it is. – José Carlos Santos Oct 05 '19 at 22:45
  • Yes there is nothing incorrect about your answer, but I just want to know why can't I state the domain as (0,∞) because that is the domain of the derivative I calculated which is 3y/2√y – harold232 Oct 05 '19 at 22:50
  • Because the method that you are using to compute the derivative of $h$ doesn't work at $0$ and therefore you deduce nothing about whether $h$ is or not differentiable at $0$. – José Carlos Santos Oct 05 '19 at 22:53
  • So basically for all cases the domain of the derivative is always the same as the domain of the original function? So for my answer, I have to rationalize the denominator to get the same domain as the original? – harold232 Oct 05 '19 at 22:56
  • I'm unsure if that is true because, look at an example like y= √x. the derivative is y' = 1/2√x. Now the domains are different – harold232 Oct 05 '19 at 22:57
  • Indeed it is not true that the domain of the derivative is always the same as the domain of the original function. But it is true in the case of your function $h$. – José Carlos Santos Oct 05 '19 at 22:59
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You haven’t even given a domain for $h$.

Here’s an answer that probably addresses your confusion:

  • The maximal domain of the formal assignment $y ↦ \frac {3y} {2 \sqrt y}$ within the real numbers is $(0..∞)$.
  • The maximal domain of the formal assignment $y ↦ \frac {3 \sqrt y} 2$ whithin real numbers is $[0..∞)$.

However, the formal assignment $y ↦ \frac {3y} {2 \sqrt y}$ defines a continuous map that can be extended in a continuous way, which can even be expressed by another formal assignment. This is probably what you are looking for. The concept behind this are removable singularities.

Also note: The maximal domains of these assignments become even larger when considering complex numbers. (However, the assignments wouldn’t give unique maps then.)

Anyway, here are the two ways to answer your original question:

  • The domain of the derivative of a function is, by definition, the same as the domain of the original function. This only works, of course, if you actually give a domain for a function and in particular one on which the function is differentiable.
  • The domain of the derivative of a function can hence be viewed as the maximal domain where the original function is differentiable. As shown by José, that would be indeed $[0..∞)$ in this case. (However, this might depend on your notion of differentiability. Some authors only consider differentiability only at interior points of the domain of a function …)
k.stm
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  • You spent enough time to write your answer, so (+1). – hamam_Abdallah Oct 05 '19 at 22:47
  • Agree but how does the "definition" override the fact that there was a radical in the denominator of the calculated derivative? I think there are cases where the domain of the derivative is different than the original function – harold232 Oct 05 '19 at 22:48
  • For example consider f(x) = √x. Domain of f(x) is [0,∞). The derivative of f(x) is f'(x)= 1 / 2√x. Notice the radical in the denominator. Now you have to exclude the 0 since division by 0 is not allowed. So the domain of f'(x) is (0,∞). Therefore in this case the domain of f and f' are not the same – harold232 Oct 05 '19 at 22:52
  • @harold232 As I said, in such cases the function is not differentiably on the entirety of its maximal domain. Then the maximal domain of the “function’s derivative” (which isn’t a well-defined notion if you don’t consider functions to come with a given domain and codomain) is the maximal domain on which the function is differentiable. – k.stm Oct 05 '19 at 22:59
  • By the way, I also suspect that you confuse the notion of differentiability with the possibility of calculating formal derivatives of expressions. Try to think of derivatives as maps $f' \colon A → B$ defined for differentiable maps $f \colon A → B$ instead of expressions (like $\frac 1 {2 \sqrt y}$) defined for “differentiable expressions” (like $\sqrt y$), whatever that may be – at least until you got things clear in your head and removed all the confusions you had. Then you can return to – or better incorporate – the vague, but more flexible notions. – k.stm Oct 05 '19 at 23:00
  • For f '(x)= 1 / 2√x, even if you rationalize the denominator, you will get f'(x)= √x / 2x, which still takes the domain (0,∞). In particular, is it just a coincidence that rationalizing the denominator for 3y / 2√y happens to change the domain to being the same as the original? Also, how would you know the domain of the derivative is always the same as the original function. Like root x gives a different domain when calculating the derivative – harold232 Oct 05 '19 at 23:06
  • Wait I now know the problem, ignore the above stuff I said. I just want to know why would I need to rationalize the denominator if it contains a radical. Why not just leave it as is. Like 1/ 2√x, no need to rationalize this – harold232 Oct 05 '19 at 23:23