I'm puzzled with the statement of differential of inverse function.
Let $f: X\to Y$ be a differentiable function.
If there exists $g:Y\to X$ s.t. $y=f(x) \Leftrightarrow x=g(y)$, then $g$ is called the inverse function of $f$.
And then, $\dfrac{dg(y)}{dy}=\dfrac{1}{\frac{df(x)}{dx}}$ holds.
I understand the statement above, but in another literature, the statement is a bit different. The statement is below.
Let $y=f(x)$ be differentiable.
If there exists the inverse function $y=g(x)$, then $\dfrac{dy}{dx}=\dfrac{1}{\frac{dx}{dy}} \ \cdots (\ast)$ holds.
I'm puzzled by this statement.
This mentions that $y=g(x)$ is inverse of $y=f(x)$, but if we write so, $f(x)=y=g(x)$ would hold but of course this doesn't hold. I cannot understand why this replaces $x=g(y)$ with $y=g(x)$.
And in $(\ast),$ I think the LHS means $\dfrac{dg(x)}{dx}$ but what does $\dfrac{dx}{dy}$ in the RHS mean ? Perhaps $\dfrac{dx}{dy}$ may mean $\dfrac{dg(y)}{dy}$ but the statement doesn't describes $x=g(y)$.
I'm confused. I'd like you to give me any help to understand this.
If $y=f(x)$ and $x=g(t)$, then the first derivative:
$\frac{dy}{dt}=\frac{df}{dx}\frac{dg}{dt}$.
If the inverse function of $y=f(x)$ is $x=g(y)$, then $t=y$ and
$\frac{dy}{dy}=\frac{df}{dx}\frac{dg}{dy}=1$,
$\frac{dg}{dy}=\frac{1}{\frac{df}{dx}}$.