Break this calculation explaining the process

If $y = \sin^{-1} x$, then by definition of the inverse sine function, we can write $x$ in terms of $y$:

$$x = \sin y$$

We differentiate both sides of $x = \sin y$ with respect to $y$:

$$\frac{dx}{dy} = \frac{d}{dy}(\sin y) = \cos y$$

We use the trigonometric identity $\sin^2 y + \cos^2 y = 1$.
From this, we can express $\cos y$ as:

$$\cos y = \sqrt{1 - \sin^2 y}$$

Since we know $x = \sin y$, we can substitute $x$ for $\sin y$:

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

Now we replace $\cos y$ in our expression for $\frac{dx}{dy}$:

$$\frac{dx}{dy} = \sqrt{1 - x^2}$$

To find $\frac{dy}{dx}$, we use the reciprocal relationship between derivatives:

$$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$$

Substitute the expression for $\frac{dx}{dy}$:

$$\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$$