Prove a functions inverse is differentiable on a domain if it is always increasing or decreasing My calculus textbook gave this theorem without a formal proof:

theorem Suppose that the domain of a function f is an open interval on which $f' (x) &gt; 0$ or on which $f '(x) &lt; 0$. Then $f$ is one-to-one, $f^{−1} (x)$ is differentiable at all values of $x$ in the range of $f$, and the derivative of $f^{−1} (x)$ is given by Formula (2). $\dfrac{dx}{dy}= \dfrac{1}{\dfrac{dy}{dx}}$
Anton, Howard. Calculus, 10th Edition (Page 428). Wiley Higher Ed. Kindle Edition.

For my class I do not need to know the proof but an understanding of what would a formal proof would look like would be helpful.
Furthermore is there a name to this proof.
I conceptually understand this proof as an increasing or decreasing function's inverse would be a function. Therefore, since the function is differentiable and it's inverse is also a function then the inverse would be differentiable. Correct me if I am wrong or if my wording makes no sense.
The fact that the inverse is a function doesn't mean it's differentiable.  For example, $f(x) = x^3$ is an increasing differentiable function on $(-\infty, \infty)$, its inverse is a function, but the inverse is not differentiable at $0$.  You need the added information that $f'$ is never $0$.  See inverse function theorem.
The fact that the inverse is a function doesn't mean it's differentiable.  For example, $f(x) = x^3$ is an increasing differentiable function on $(-\infty, \infty)$, its inverse is a function, but the inverse is not differentiable at $0$.  You need the added information that $f'$ is never $0$.  See inverse function theorem.