Derivative of $\arcsin(x)=\frac{1}{\cos(x)}$ . 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 $\arcsin(x)\neq\frac{1}{\cos(x)}$?
 A: 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}} $
A: They are reciprocal in the sense that, if the derivative of $\sin x$ is $\cos x$, then at the $y$ where $\arcsin y = x$, the derivative of $\arcsin y$ (w.r.t $y$) is $\frac1{\cos x}$.
Then to represent that derivative using $y$, one has to simplify $\frac{1}{\cos (\arcsin y)}$.
In fact, $\arcsin y = x$ means $y = \sin x$, so
$$\frac{d}{dy}\arcsin y = \frac{1}{\sqrt{1-y^2}} = \frac{1}{\sqrt{1-\sin^2 x}} = \frac{1}{|\cos x|} = \frac{1}{\cos x}$$
Assuming $x$ is in the range of $\arcsin$, and simultaneously when $\cos x\ne 0$ so $\arcsin y$ is differentiable w.r.t $y$, i.e. $x\in \left(-\frac{\pi}{2}, \frac\pi2\right)$.
A: $$y=\arcsin(x)\;, x= \sin y\;$$
Differentiate
$$ dx= \cos y \; dy $$
$$\dfrac{dy}{dx}=\dfrac{1}{\cos y}=\dfrac{1}{\sqrt{1-\sin^2y}}=\dfrac{1}{\sqrt{1-x^2}}. $$
