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?
Let $\arcsin(x+h)=u$ and $\arcsin x=v$

$\displaystyle \implies (i) x+h=\sin u, x=\sin v$

and $\displaystyle (ii) -\frac\pi2\le u,v\le \frac\pi2$ (using the definition of principal value)

$\displaystyle\implies \cos u,\cos v\ge0$

$$\displaystyle\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin x}h=\lim_{u\to v}\frac{u-v}{\sin u-\sin v}$$

$$\displaystyle=\lim_{u\to v}\frac{u-v}{2\sin\frac{u-v}2\cos\frac{u+v}2}$$

$$\displaystyle=\lim_{u\to v}\frac{\frac{u-v}2}{\sin\frac{u-v}2} \cdot\frac1{\lim_{u\to v}\cos\frac{u+v}2}$$

$$\displaystyle=1\cdot\frac1{\cos v}=+\frac1{\sqrt{1-x^2}}$$
Can be prove by the following:

$y=arc\sin x\to y'=y'_x=\frac{1}{x'_y}=\frac{1}{cosy}=\frac{1}{\sqrt{1-sin^2y}}=\frac{1}{\sqrt{1-x^2}}$