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Let $F:[0,1]\rightarrow \mathbb{R} $ $$F(x)=\int _{\sqrt x}^{1}\arcsin(t^2) \,dt$$

Is $F$ differentiable?


The function $f(t)=\arcsin(t^2)$ is continuous on $[0,1]$ so is integrable on $[0,1]$ and $$ \frac{d}{dx}\sqrt x=\frac{1}{2\sqrt x}$$

So $F$ is differentiable and by the F.T.C

$$ F'(x)=-\frac{\arcsin(x)}{2\sqrt x}$$

Is correct my answer?

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    $\begingroup$ What about $F’(0)$? $\endgroup$ – ziggurism Jun 2 '18 at 18:17
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    $\begingroup$ What does $\;\int_{\sqrt x}^1=\arcsin t^2\;$ mean?? Twice you wrote this so I am guessing it isn't a typo... $\endgroup$ – DonAntonio Jun 2 '18 at 18:21
  • $\begingroup$ @DonAntonio It is a typo, sorry $\endgroup$ – B. David Jun 2 '18 at 18:28
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You have clearly a problem when $x=0$. However, it is easy to solve:$$\lim_{x\to0}F'(x)=\lim_{x\to0}\sqrt x\frac{\arcsin x} x=\sqrt0\times1=0.$$It is well-known that it follows from this that $F'(0)=0$.

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  • $\begingroup$ Thank you so much for your help. So can I say $F(x)$ is differentiable on (0,1)? $\endgroup$ – B. David Jun 2 '18 at 18:47
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    $\begingroup$ @B.David You can say that it is differentiable on $[0,1)$. $\endgroup$ – José Carlos Santos Jun 2 '18 at 18:50
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    $\begingroup$ Well, in fact differentiable on the right at $\;x=0\;$ since $\;F\;$ is only defined on the right of zero. $\endgroup$ – DonAntonio Jun 2 '18 at 18:53
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You established that the derivative for $x\ne 0$. To determine whether $F'(0^+)$ exists, we analyze the limit

$$\begin{align} \lim_{h\to 0^+}\frac{\int_{\sqrt h}^1\arcsin(x^2)\,dx-\int_0^1 \arcsin(x^2)\,dx}{h}&=-\lim_{h\to 0^+}\frac1h \int_0^{\sqrt h}\arcsin(x^2)\,dx\\\\ &=-\lim_{h\to 0^+}\frac1{2h} \int_0^{h}\frac{\arcsin(x)}{\sqrt x}\,dx\\\\ &=-\frac12 \lim_{h\to 0^+}\frac{\arcsin(h)}{\sqrt h}\\\\ &=0 \end{align}$$

Hence, $F'(0^+)=0$ where the derivative is the right-sided derivative. And we are done!

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An idea for you: since $\;\arcsin t^2\;$ is continuous on the integration interval whatever $\;x\;$ is, there exists a differentiable $\;G\;$ s.t.

$$F(x):=\int_{\sqrt x}^1\arcsin t^2\;dt=\left.G(t)\right|_{\sqrt x}^1=G(1)-G(\sqrt x)\implies F'(x)=$$

$$=-G'(\sqrt x)\cdot\frac1{2\sqrt x}=-\frac{\arcsin x}{2\sqrt x}$$

and among other things, $\;F(0)=G(1)-G(0)\;$ , and now:

$$F'(0)=\lim_{x\to0}\frac{F(x)-F(0)}x=\lim_{x\to0}\frac{G(1)-G(\sqrt x)-G(1)+G(0)}x=$$

$$=-\lim_{x\to0}\frac{G(\sqrt x)-G(0)}x=-\left.G'(\sqrt x)\right|_{x=0}=-G'(0)=-\arcsin 0=0$$

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