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?