Calculate the derivative and find its domain: $\;f(x)= \sqrt{\ln(x)+2}$

Question

I calculated the derivative as $$f'(x) = \frac{1}{2x \sqrt{2+\ln x}}$$

How do I find out the domain?

score 4 · Answer 1 · answered Oct 29 '14 at 17:58

4

To make sure that the denominator $\not = 0$, we need to require

$$\{x:\ln x+2\gt 0\}=(e^{-2},\infty)$$

answered Oct 29 '14 at 17:58

mike

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I would require $x>0 \cap {\ln x + 2 >0}$ (initially excluding $x=0$), instead. The result does not change, though. – Avitus Oct 29 '14 at 18:01
$x\not = 0 \cap {\ln x + 2 >0}$ is better. Thanks! – mike Oct 29 '14 at 18:03
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...and $x>0$ due to the presence of $\ln x$. Your result is correct (+1) – Avitus Oct 29 '14 at 18:25

score 1 · Answer 2 · answered Oct 29 '14 at 17:55

1

The domain is the set $\{x:\ln x+2\ge 0\}=[e^{-2},\infty)$

answered Oct 29 '14 at 17:55

Why? can you show your steps – user3924310 Oct 29 '14 at 17:58
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At $x = e^{-2}$, the derivative is not defined, in case the OP was looking for the domain of $f'(x)$. – amWhy Oct 29 '14 at 18:00
@user3924310 I am assuming that you want the domain of the function and I have also assumed that the function is real valued. Otherwise, the domain is the whole $\mathbb{R}$ – Samrat Mukhopadhyay Oct 29 '14 at 18:01
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If the OP is asking for the domain of the derivative then the domain is $(e^{-2},\infty)$. Basically by domain we mean the set of values of $x$ (in $\mathbb{R}?$) for which the function does not "blow up" to $\pm \infty$. – Samrat Mukhopadhyay Oct 29 '14 at 18:05

2 Answers2