It can be expressed in two different forms:

f(x) = g^I(h(x)) * h^I(x)

Or in Leibniz notation as:
dy/dx = (dy/du)(du/dx)

To get the derivative of a composite function you do this:

f(x) = sin(x³ + 5)

We split this into two separate functions:
g(x) = sin(x) and h(x) = x³ + 5

We then find the derivative of each function:
g^I(x) = cos(x) and h^I(x) = 3x²

Then simply plug them into the formula:
f^I(x) = g^I(h(x)) * h^I(x)
f^I(x) = cos(x³ + 5) * 3x²

Note: The chain rule can be applied to composite functions that contain more than just two separate functions like so:

f(x) = (sin(x³))⁴

We split this into three separate functions:
g(x) = x⁴ and h(x) = sin(x) and i(x) = x³

We then again find the derivatives of these functions:
g^I(x) = 4x³
h^I(x) = cos(x)
i^I(x) = 3x²

We can then express f^I(x) as:
f^I(x) = g^I(h(i(x))) * h^I(i(x)) * i^I(x)

Which when we plug the numbers in looks like:
f^I(x) = 4(sin(x³))³ * cos(x³) * 3x²

I hate to be nitpicky but izubachi started it.

What you illustrate is not so much that the chain rule can be applied to non-composite functions but rather that all functions can be seen as composites. Which is sort of common sense.