Derivative of $\arcsin(x)=\frac{1}{\cos(x)}$ .

Question

My intuition says that since $\sin(x)$ and $\arcsin(x)$ are inverse of each other, their derivatives must be reciprocal. I have previously proved the fact that $\arcsin(x)'=\frac{1}{\sqrt{1-x^2}}$ but it doesn't match my thought about being reciprocal and also observation.

Here is the illustration showing the slope is probably reciprocal of each other:

The question is why is the idea wrong in this case i.e $\arcsin(x)\neq\frac{1}{\cos(x)}$?

Answer 1

If $ y = \arcsin(x) $

Then,

$ x = \sin(y) $

Using implicit differentiation,

$ 1 = \cos(y) y' $

Hence,

$ y' = \dfrac{1}{\cos(y)} $

But $\cos(y) = \sqrt{1 - \sin(y)^2} = \sqrt{1 - x^2}$

Therefore,

$y' = \dfrac{1}{\sqrt{1 - x^2}} $