A method used to solve

derivatives of

composite functions such as: f(x) = sin(x

^{3} + 5)

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(x^{3} + 5)

We split this into two separate functions:

g(x) = sin(x) and h(x) = x^{3} + 5

We then find the derivative of each function:

g^{I}(x) = cos(x) and h^{I}(x) = 3x^{2}

Then simply plug them into the formula:

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

f^{I}(x) = cos(x^{3} + 5) * 3x^{2}

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

f(x) = (sin(x^{3}))^{4}

We split this into three separate functions:

g(x) = x^{4} and h(x) = sin(x) and i(x) = x^{3}

We then again find the derivatives of these functions:

g^{I}(x) = 4x^{3}

h^{I}(x) = cos(x)

i^{I}(x) = 3x^{2}

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(x^{3}))^{3} * cos(x^{3}) * 3x^{2}

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.