Is taking the reciprocal of a differentiable strictly rigorous?

Question

I've read a proof of $\frac{d}{dx}(\ln x)=\frac{1}{x}$ in which the reciprocal of $\frac{dx}{dy}$ is taken and was wondering if this is strictly rigorous. The proof was: $$y=\ln x$$ $$e^y=x$$ then take the derivative with respect to $y$, so $$e^y=\frac{dx}{dy}$$ then take the reciprocal of the differentiable to give $$\frac{1}{e^y}=\frac{dy}{dx}$$ and since $e^y=x$, $$\frac{dy}{dx}=\frac{1}{x}$$

the question is simply: is taking the reciprocal of a differentiable rigorous? I know considering it as a variable to add or subtract to different sides of an equation, but I (obviously) am not sure about this case. Any help is appreciated.

Answer 1

This is a consequence of the chain rule. For a function composition $F=f \circ g$, the rule states $$F'(x)=f'(g(x))g'(x)$$. In other words, if you denote $y=g(x)$ and $z=f(y)$, then $$\frac{dz}{dx}=\frac{dz}{dy}\frac{dy}{dx}$$ Now if $g$ is an invertible function and $f$ is its inverse, then $$F=g^{-1}\circ g$$ is just the identity $F(x)=x$. In other words, $z=x$. Then $$\frac{dz}{dx}=1=\frac{dx}{dy}\frac{dy}{dx}$$ or $$\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}$$ That's why you were okay to do it.

Answer 2

I cannot comment, yet. Therefore, I write an answer. It is not really rigorous as presented here. But if you use the right theorem (https://en.wikipedia.org/wiki/Inverse_function_theorem) it works.