Inverse Functions and proving the product of derivatives is 1

Question

Write down the derivative of the function y=x^3 - 1: 3x^2

Make x the subject and hence find dx/dy: 1/3(y+1)^-2/3

Show that dy/dx * dx/dy = 1

How does the x and y possibly cancel out?

It seems like a very straightforward question but what do i do?

Answer 1

Well, we have

$$\frac{dx}{dy}=\frac13(y+1)^{-2/3}=\frac13((x^3-1)+1)^{-2/3}=\frac13x^{-2}$$

Then can you show the product is one?

Answer 2

Lets prove that $\frac{dy}{dx}\cdot \frac{dx}{dy}$

Let $f$ an invertible function defined as $f: X\to Y$ and its inverse $f^{-1}:Y \to X$ and $x \in X, y \in Y$.

Also, let $f(x) = y \implies f^{-1}(y)=x$.

Using implicit derivation on $f^{-1}(y)=x$ respect to $x$ you get:

$\frac{d}{dx}f^{-1}(y) = 1 \xrightarrow{\text{Applyng chain rule in LHS}} \frac{d}{dy}f^{-1} \cdot \frac{d}{dx}y = 1\xrightarrow{\text{Rewritting terms}} \frac{dx}{dy} \cdot \frac{dy}{dx} = 1$

$ \square$