question about the limit $\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin(x)}{h}$ Because $\sin'(x)=\cos(x)$ we can prove that $\arcsin'(x)=\frac{1}{\sqrt{1-x^2}}$. but, by definition we have
$$\arcsin'(x)=\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin(x)}{h}\tag{1}$$
therefore,
$$\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin(x)}{h}=\frac{1}{\sqrt{1-x^2}}\tag{2}$$
My question: Can we prove $(2)$ without using $(1)$? How?

Here is one way forward.  Note that $\sin (\arcsin (x))=x$ and $\cos (\arcsin(x))=\sqrt{1-x^2}$ for $|x|\le 1$.  Therefore, we have for $|x|\le 1$
$$\frac{\arcsin(x+h)-\arcsin(x)}{h}=\frac{\arcsin\left((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}\right)}{h} \tag 1$$
Now, in THIS ANSWER, I showed using only basic inequalities that for $|x|<1$
$$\left|\frac{x}{\sqrt{1-x^2}}\right|\ge |\arcsin (x)|\ge |x| \tag 2$$
Using the right-hand side inequality of $(2)$ in $(1)$ yields
$$\begin{align}
\left|\frac{\arcsin\left((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}\right)}{h}\right|&\ge \left|\frac{(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}}{h}\right|\\\\
&=\frac{(x+h)^2(1-x^2)-x^2(1-(x+h)^2)}{h\left((x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}\right)}\\\\
&=\frac{2x}{(x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}} \tag 3
\end{align}
$$
And using the left-hand side inequality of $(2)$ in $(1)$ yields
$$\begin{align}
\left|\frac{\arcsin\left((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}\right)}{h}\right|&\le \frac{\frac{2x}{(x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}}}{\sqrt{1-\left((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}\right)^2}} \tag 4
\end{align}
$$
Using the Squeeze Theorem, we obtain from $(3)$ and $(4)$
$$\lim_{h\to 0}\frac{\arcsin(x+h)-\arcsin(x)}{h}=\frac{1}{\sqrt{1-x^2}}$$
