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I have the following function:

$f(x) = \arcsin{(\sqrt x)}$

I've caculated the derivative to:

$f'(x)=\frac{1}{2 \sqrt{x} \cdot \sqrt{1-x}}$

And the domain of $f(x)$ to $[0, 1]$

And the domain of $f'(x)$ to $(0, 1)$

I want to determine for which $x$ the derivative exists but I'm not really sure if I should use the domain of the original function or of the derivative of the function to know where the derivative exists?

Integrand
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freya
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1 Answers1

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The domain of the derivative is what matters here – it's not defined at the endpoints of the interval, so the derivative is defined on $(0,1)$. (However, one-sided derivatives exist at $0$ and $1$.)

Parcly Taxel
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  • Thank you!! So it should be enough to calculate the domain of the derivative to prove it exists? – freya Sep 11 '20 at 16:47
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    @freya You'd also have to take out points where the original $f$ isn't defined. But in this case $f(x)$ is well-defined on $(0,1)$, so nothing needs to be plucked out. – Parcly Taxel Sep 11 '20 at 16:49