Is the domain of a derivative always the same as the domain of the original function?

Question

I have seen similar if not the same questions, but none of them fully answer this question.

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For my question, I want to know why I would need to rationalize the denominator of $\frac {3y} {2 \sqrt y}$ to $\frac {3 \sqrt y} 2$, when writing out the derivative for $h(y) = y\sqrt y$. In another similar case for $\frac {1} {2 \sqrt x}$ to $\frac {\sqrt x} {2 \ x}$, for the derivative of y = √x, there is no need to rationalize the denominator of its derivative; in other words, the domain of its derivative is different from the original function, regardless if I rationalize the denomiantor or not. So what is the purpose of rationalizing the denominator?

Note:

-The domain of the derivative of h(y) is [0,∞) when leaving it as $\frac {3y} {2 \sqrt y}$

• The domain of the derivatibe of h(y) is (0,∞) when rationalizing the denominator to $\frac {3 \sqrt y} 2$

• The domain of h(y) is [0,∞)

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For those who still don't understand my question:

For h'(y), why would I need to rationalize the denominator? How would I know that the domains of h and h' are the same in the first place? From initial calculation of the derivative, it gives $\frac {3y} {2 \sqrt y}$, where x at 0 is undefined

Answer 1

For $y = \sqrt x$, it is clear that the derivative is not defined at x=0 because y'= $\frac {1} {2 \sqrt x}$ will result in $\frac {1} {0}$, and to double check $\frac {\sqrt x} {2 \ x}$ still makes the derivative undefined x=0 resulting in $\frac {0} {0}$. So in either case, you have a 0 in the denominator, in other words resulting in division by 0, which makes the function undefined at x = 0.

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For h(y), it is a different case. You calculate h'(y) as:

$\frac {3y} {2 \sqrt y}$ or $\frac {3 \sqrt y} 2$

For $\frac {3y} {2 \sqrt y}$, you get a $\frac {0} {0}$ case, but when rationalizing the denominator to $\frac {3 \sqrt y} 2$, you get $\frac {0} {2}$, which is 0, so it is defined at 0. So in brief, from my way to look at it this is to view the step of "rationalzing the denomiantor" as a second check as to whether the derivative is really undefined at x = 0.

• Thus, the domain for h'(y) is [0,∞), as x at 0 of the derivative IS defined, so include it in the domain.

Answer 2

I'm not sure I understand your question, but it contains a couple of statements that appear to me to be incorrect. The function $\ h(y)=y\sqrt{y}=y^\frac{3}{2}\ $ is defined for all non-negative $\ y\ $, so its domain is $\ [0,\infty)\ $, not $\ (0,\infty)\ $ (unless, for some reason you want to restrict it artificially to the latter interval).

The function $\ h\ $ is differentiable on $\ (0,\infty)\ $, but not at $\ y=0\ $. While it does have a right derivative there, it has no left derivative, because $\ h(y)\ $ is undefined for $\ y < 0\ $.

Both of the expressions $\ \frac{3y}{2\sqrt{y}}\ $ and $\ \frac{3\sqrt{y}}{2}\ $ are correct ways to write the derivative of $\ h\ $, as long as you keep in mind that they're only valid for $\ y>0\ $. While $\ \frac{3y}{2\sqrt{y}}\ $ is undefined at $\ y=0\ $, this isn't a problem because the derivative doesn't exist there anyway. Also, while $\ \frac{3\sqrt{y}}{2}\ $ is a little simpler, and seems to me to be a more natural way to write the derivative, it is certainly not mandatory for you to write it that way.

Reply to query from OP in comments below

The difference between the two cases is that, in the first, one of the two equivalent expressions (namely $\ \frac{1}{\sqrt{x}}\ $) has a non-zero numerator and a zero denominator, so it's going to remain undefined at $\ x=0\ $ no matter how you write it.

In the second case, however, one of the two not quite equivalent expressions (namely $\ \frac{3\sqrt{y}}{2}\ $) has a zero numerator and non-zero denominator, and therefore has a well-defined value of $\ 0\ $ at $\ y=0\ $. But if you then multiply it by any expression of the form $\ \frac{f(y)}{f(y)}\ $ where $\ f(0)=0\ $ (and, in particular, by $\ \frac{\sqrt{y}}{\sqrt{y}}\ $), the resulting expression will be undefined at $\ y=0\ $, and therefore have a different domain from the original.