Is $F(x)=\int _{\sqrt x}^{1}\arcsin(t^2) \,dt$ differentiable? 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?
 A: 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 $F'(0)=0$.
A: 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!
A: 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:
$$F'(0)=\lim_{x\to0}\frac{F(x)-F(0)}x=\lim_{x\to0}\frac{G(1)-G(\sqrt x)-G(1)+G(0)}x=$$
$$=-\lim_{x\to0}\frac{G(\sqrt x)-G(0)}x=-\left.G'(\sqrt x)\right|_{x=0}=-G'(0)=-\arcsin 0=0$$
