Extra explanation

The formula for the derivative of the inverse function seems to appear out of nowhere, but we can prove it by the use of the property of the inverse function and the chain rule.

We use:

1. $x(y(x)) = x$,
2. The derivative of $u(v(x))$ is equal to $u'(v(x))v'(x)$.

One could consider $x(y(x))$ as a composite function with $v(x) = y(x)$ and $u(v) = x(y)$. The derivative of $x(y(x))$ is then
$$
\begin{align}
v'(x) &= y'(x)\\
u'(v) &= x'(y)\\
\tfrac{d}{dx} x(y(x)) &= u'(v(x))v'(x) = x'(y(x)) \cdot y'(x).
\end{align}
$$
We use the property of the inverse function and find the derivative of both the left and the right-hand side:
$$
\begin{align}
x(y(x)) &= x\\
\tfrac{d}{dx} x(y(x)) &= \tfrac{d}{dx} x\\
x'(y(x)) \cdot y'(x) &= 1\\
x'(y(x)) &= \dfrac{1}{y'(x)}\\
x'(y(x(y)) &= \dfrac{1}{y'(x(y))}\\
x'(y) &= \dfrac{1}{y'(x)} \qquad \text{with} \qquad x=x(y).
\end{align}
$$
At the second last step we use $x = x(y)$ and at the final step the property of the inverse function, being that $y(x(y)) = y$.