How to write $(f^{-1})'(b)=\frac{1}{f'\left(f^{-1}(b)\right)}$ in Leibnizian notation? On Calculus by Spivak question 12-17 we are asked to rewrite this equation in  Leibnizian notation 

$$(f^{-1})'(b)=\frac{1}{f'\big(f^{-1}(b)\big)}\text{.}$$

The answer book says that it should be written 

$$\dfrac{dx}{dy}=\frac{1}{\dfrac{dy}{dx}} . \tag{1}$$

But, wouldn't this be the translation for 
$$(f^{-1})'(x)=\frac{1}{f'(x)} ? \tag{2}$$
How is (1) the correct translation? If (1) is the correct translation, how would one write (2) in Leibnizian notation?
 A: Let
$$y(x) = f(x) \tag{1}\label{eq1A}$$
Using $b$ as a replacement for $y$ results in
$$x(y) = f^{-1}(y) = f^{-1}(b) \tag{2}\label{eq2A}$$
Substituting \eqref{eq2A} into \eqref{eq1A} gives
$$y(x) = y(f^{-1}(y)) = y(f^{-1}(b)) \tag{3}\label{eq3A}$$
Remember the "$'$" symbol used with a function means to differentiate that function wrt its argument. Thus, you get from \eqref{eq2A}
$$(f^{-1})'(b) = (f^{-1})'(y) = x'(y) = \frac{dx}{dy} \tag{4}\label{eq4A}$$
plus from \eqref{eq1A} and \eqref{eq3A} you get
$$f'(f^{-1}(b)) = f'(x) = y'(x) = \frac{dy}{dx} \tag{5}\label{eq5A}$$
This means you have
$$(f^{-1})'(b)=\frac{1}{f'\big(f^{-1}(b)\big)} \implies \dfrac{dx}{dy}=\frac{1}{\dfrac{dy}{dx}} \tag{6}\label{eq6A}$$
Regarding your statement of
$$(f^{-1})'(x)=\frac{1}{f'(x)} \tag{7}\label{eq7A}$$
note you're using the same argument, i.e., $x$ for both a function and its inverse in the same expression. This doesn't really work. Nonetheless, if you did try this to implement this in Leibnizian notation, it might be something like
$$\frac{dx}{dx} = \frac{1}{\dfrac{dy}{dx}} \tag{8}\label{eq8A}$$
