I have recently come across two formula's that I am unfamiliar with and would like to know if they are both aspects of the same thing:
$$\color{purple}{\cfrac{1}{f^{\prime}(a)}=f(a)(f^{-1})^{\prime}}\tag{1}$$
$$\rho_x (x)=\rho_\alpha(\alpha)\left|\frac{\mathrm{d}x}{\mathrm{d}\alpha}\right|^{-1}\tag{2}$$
$(1)$ is the Inverse function theorem and $(\mathrm{2})$ is the formula that relates probability densities for random variables $x$, $\alpha$.
In my previous post I was shown that the Inverse function theorem $(1)$ was needed to prove $(2)$.
However, is it true that
$$\underbrace{\rho_x (x)}_{=\color{purple}{\cfrac{1}{f^{\prime}(a)}}}=\underbrace{\rho_\alpha(\alpha)}_{=\color{purple}{\displaystyle f(a)}}\overbrace{\left|\frac{\mathrm{d}x}{\mathrm{d}\alpha}\right|^{-1}}^{=\color{purple}{\displaystyle (f^{-1})^{\prime}}}$$
Can someone please explain whether the above statement is true?
$$\frac{\mathrm{d}\rho}{\mathrm{d}x}(x)=\rho_x (x)=\underbrace{\rho_ \alpha(\alpha)}_{=\color{purple}{\displaystyle \frac{d\rho}{d\alpha}(\alpha)}}\overbrace{\left|\frac{\mathrm{d}x}{\mathrm{d}\alpha}\right|^{-1}}^{=\color{purple}{\displaystyle \left|\frac{d\alpha}{dx}\right|}}= \frac{\mathrm{d}\rho}{\mathrm{d}\alpha}(\alpha)\left|\frac{\mathrm{d}\alpha}{\mathrm{d}x}\right| $$
without the absolute value the above is just the chain rule $$\frac{df}{dx}=\frac{df}{dy}\frac{dy}{dx}$$
also note that $$\left(\frac{dx}{d\alpha}\right)^{-1}=\frac{1}{\frac{d\alpha}{dx}}=\frac{d\alpha}{dx}$$