what is wrong with the following reasoning?
$$\sin(\sin^{-1}(x))=x$$
This is of the form $f(g(x))=h(x)$. Then, by implicit differentiation:
$$(f(g(x)))'=f'(g(x))\cdot g'(x)=h'(x) $$
Or in this case:
$$\cos(\sin^{-1}(x))\frac{d(\sin^{-1}(x))}{dx} = 1$$
Then $$\frac{d(\sin^{-1}(x))}{dx} = \frac{1}{\cos(\sin^{-1}(x))}$$
However I do not think this is correct. I wonder if I made a conceptual mistake in the implicit differentiation. Perhaps there are some extra steps needed to get to the result $$\frac{d(\sin^{-1}(x))}{dx} = \frac{1}{\sqrt{1-x^2}}$$
If anyone could help me out that would be awesome! Thanks in advance.