Proof of Derivative of sin inverse $ sin^{-1} (x) $

Question

Here are the first few lines of the proof:

$$ y = \sin^{-1} (x)$$ $$ x = \sin (y)$$ $$ \cos y(dy/dx) = 1$$ $$\vdots$$

Why is the $(dy/dx)$ retained in the third step? Isn't the derivative of sin $y = \cos x$ and hence shouldn't $(dy/dx)$ be omitted in the third step?

PS: I found this on several sources and so it isn't a typo.

Answer 1

It's chain rule.

You're treating $x$ as a function of $y$, $x = f(y)$. In this case, $f(y) = \sin y$.

So $\displaystyle \frac{d f(y)}{dx} = \frac{df(y)}{dy}\cdot\frac{dy}{dx} = \cos y \frac{dy}{dx}$

Answer 2

The chain rule states $$ \frac{du(v(t))}{dx} = \frac{du}{dv}\frac{dv}{dx} $$ When differentiating, $$\begin{align*} \frac{d}{dx}x &= \frac{d}{dx} \sin y\\ 1 &= \frac{d\sin y}{dy}\frac{dy}{dx} \\ 1 &= \cos y\,\frac{dy}{dx} \\ \end{align*}$$ In this case, $u = \sin y, v = y$.