First we are going to restrict the domain of $sin$ to be monotone, $ [ - \frac{\pi}{2}, \frac{\pi}{2}]$; Let’s also call the inverse of $sin$, $arcsin$
$\therefore sin(arcsin(x)) = x, \text{ for } \in [-1, 1]$
$\therefore arcsin(sin(x)) = x, \text{ for } \in [- \frac{\pi}{2}, \frac{\pi}{2}]$
Derivative of $arcsin$? Use the inverse derivative!
$(arcsin)’(x) = \frac{1}{sin’(arcsin(x))} = \frac{1}{cos(arcsin(x))}$
Note: $\cos^2 x + \sin^2 x = 1 \Leftrightarrow \cos^2 \arcsin x + \sin^2 \arcsin x = 1 \Leftrightarrow$
$(\cos \arcsin x)^2 + x^2 = 1 \Leftrightarrow \cos \arcsin x = \sqrt{ 1 - x^2} \Leftrightarrow$
$\therefore \frac{1}{\cos \arcsin x} = \frac{1}{\sqrt{1 - x^2}}$
Conclusion: $\arcsin(x)$ is a differentiable (defined and continuos) on $[-1, 1]$ and $arcsin’(x) = \frac{1}{\sqrt{x^2 - 1}}$
we are similarly going to restrict the domain to $[0, \pi]$
Let $ \cos^{-1} = \arccos$
$\therefore arccos(cos(x)) = x, \text{ for } \in [ 0 , \pi]$
$\therefore cos(arccos(x)) = x, \text{ for } \in [-1, 1]$
Derivative of $\arccos$
$(\arccos)’(x) = \frac{1}{\cos ’ (\arccos (x))} = \frac{1}{\cos ’ (\arccos (x))}$
Note: we can use our same logic before
*$\sin^2 (\arccos(x)) + \cos^2 (\arccos(x)) = 1 \Leftrightarrow (\sin (\arccos(x)))^2 + x^2 = 1$ \
$\frac{1}{\cos ’ (\arccos (x))} = \frac{1}{\sqrt{1 - x^2}}$
$\therefore (\arccos)’(x) = \frac{1}{\sqrt{1 - x^2}}$
you get the story
we restrict it by $[ - \frac{\pi}{2}, \frac{\pi}{2}]$
we call it $\arctan$
we get the derivative: $ (\arctan)’(x) = \frac{1}{1 + x^2}$