# Math Help - Domain sizes

1. ## Domain sizes

I came across this question in one of my old textbooks:

If the function y = f(x) has domain D and its derivative y' = f'(x) has domain D', what can one say about the relative sizes of D and D'?

After a bit of thought, I found 3 examples having D < D', D = D', and D > D':

$y = x^x \rightarrow y' = x^x (\ln x + 1)$
$y = x^2 \rightarrow y' = 2x$
$y = \ln x \rightarrow y' = \frac{1}{x}$

So, is the answer, it depends? Or am I missing something?

-Scott

2. Originally Posted by ScottO
I came across this question in one of my old textbooks:

If the function y = f(x) has domain D and its derivative y' = f'(x) has domain D', what can one say about the relative sizes of D and D'?

After a bit of thought, I found 3 examples having D < D', D = D', and D > D':

$y = x^x \rightarrow y' = x^x (\ln x + 1)$
$y = x^2 \rightarrow y' = 2x$
$y = \ln x \rightarrow y' = \frac{1}{x}$

So, is the answer, it depends? Or am I missing something?

-Scott
Hello,

1. You've found three different possible relations between the domain of a function and the domain of the first derivative of this function. But the first case you mention is impossible:

If $D(f)\subset D(f')$ then there exist at least one x for which f(x) doesn't exist but the derivative. Or in plain words: That can't be.
The only relation between these two sets is: $D(f) \subseteq D(f')$

i) $y = x^x ~,\ x\in\mathbb{R}_0^+$ because the limit exist: $\lim_{x\to 0+}\left(x^x\right)=1$.
But: $y' = x^x (\ln x + 1)~,\ x\in\mathbb{R}^+$ because ln(0) is not defined. So $D(f')\subset D(f)$
ii) $y = x^2 \rightarrow y' = 2x$ Obviously $D(f) = D(f')$
iii) $y = \ln x \rightarrow y' = \frac{1}{x}$ . The ln-function is defined for positive real numbers only and therefore the domain of the first derivation can't be greater than the domain of the function (see above). Thus $D(f) = D(f')$