I'm a bit confused about how the inverse trig functions are differentiated. From a website:
We have the following relationship between the inverse sine function and the sine function.
sin(sin−1x)=xsin−1(sinx)=xsin(sin−1x)=xsin−1(sinx)=x
In other words they are inverses of each other. This means that we can use the fact above to find the derivative of inverse sine. Let’s start with,
f(x)=sinxg(x)=sin−1xf(x)=sinxg(x)=sin−1x
Then,
g′(x)=1f′(g(x))=1cos(sin−1x)
This is not a very useful formula. Let’s see if we can get a better formula. [..]
Why is the author saying this is not a very useful formula and we need to continue further? It looks like the derivative of g′(x) is expressed in terms of x which is what we want and we applied the inverse function rule correctly.
Is there a reason for continuing until we arrive at:
ddx(sin−1x)=1√1−x2
