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?
 A: 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)
$\displaystyle\implies \cos u,\cos v\ge0$
$$\displaystyle\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin x}h=\lim_{u\to v}\frac{u-v}{\sin u-\sin v}$$
$$\displaystyle=\lim_{u\to v}\frac{u-v}{2\sin\frac{u-v}2\cos\frac{u+v}2}$$
$$\displaystyle=\lim_{u\to v}\frac{\frac{u-v}2}{\sin\frac{u-v}2} \cdot\frac1{\lim_{u\to v}\cos\frac{u+v}2}$$
$$\displaystyle=1\cdot\frac1{\cos v}=+\frac1{\sqrt{1-x^2}}$$
A: $$\text{Using }\arcsin p-\arcsin q=\arcsin(p\sqrt{1-q^2}-q\sqrt{1-p^2}),$$
$$\arcsin(x+h)-\arcsin x=\arcsin((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2})$$
$$\displaystyle\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin x}h $$
$$=\lim_{h\to0}\frac{\arcsin((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2})}h$$
$$=\lim_{h\to0}\frac{\arcsin((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2})}{(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}}\cdot\lim_{h\to0}\frac{(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}}h$$
Putting $\displaystyle\arcsin((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2})=v$
$\displaystyle\implies(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}=\sin v$ in the first limit,
$$\lim_{h\to0}\frac{\arcsin((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2})}{(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}}=\lim_{v\to0}\frac v{\sin v}=1$$
$$\text{Again for the second limit, }\lim_{h\to0}\frac{(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}}h$$
$$=\lim_{h\to0}\frac{(x+h)^2(1-x^2)-x^2(1-(x+h)^2)}{h\{(x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}\}}$$
$$=\lim_{h\to0}\frac{(x+h)^2-x^2}h\cdot\lim_{h\to0}\frac1{(x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}}$$
$$=\lim_{h\to0}\frac{h(2x+h)}h\cdot\frac1{2x\sqrt{1-x^2}}$$
$$=\lim_{h\to0}(2x+h)\cdot\frac1{2x\sqrt{1-x^2}}\text{ as } h\ne0\text{ as }  h\to0$$
$$=2x\cdot\frac1{2x\sqrt{1-x^2}}=\frac{1}{\sqrt{1-x^2}}$$
A: 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 $(4)$
$$\lim_{h\to 0}\frac{\arcsin(x+h)-\arcsin(x)}{h}=\frac{1}{\sqrt{1-x^2}}$$
A: Can be prove by the following:
$y=arc\sin x\to y'=y'_x=\frac{1}{x'_y}=\frac{1}{cosy}=\frac{1}{\sqrt{1-sin^2y}}=\frac{1}{\sqrt{1-x^2}}$
