# Derivatives of Composite Functions - Chain Rule

#### notes

To find the derivative of  f, |
where f(x) = (2x + 1)^3
One way is to expand (2x + 1)3 using binomial theorem and find the derivative as a polynomial function as illustrated below.
d/(dx)f(x) = d/(dx) [(2x + 1)^3]

=d/(dx) (8x^3 + 12x^2 + 6x + 1)

= 24x^2 + 24x + 6

= 6 (2x + 1)^2
Now, observe that

f(x) = (h o g) (x)

where g(x) = 2x + 1 and h(x) = x^3.

Put t = g(x) = 2x + 1. Then f(x) = h(t) = t^3. Thus

(df)/(dx) = 6(2x + 1)^2 = 3(2x + 1)^2 . 2 = 3t^2 . 2 = (dh)/(dt) . (dt)/(dx)
The advantage with such observation is that it simplifies the calculation in finding the derivative of, say, (2x + 1)^100. We may formalise this observation in the following theorem called the chain rule.

#### theorem

Theorem: (Chain Rule) Let f be a real valued function which is a composite of two functions u and v; i.e., f = v o u. Suppose t = u(x) and if both (dt)/(dx) and (dv)/(dt) exist , we have
(df)/(dx) = (dv)/(dt) . (dt)/dx
We skip the proof of this theorem. Chain rule may be extended as follows. Suppose f is a real valued function which is a composite of three functions u, v and w; i.e., f = (w o u) o v. If t = v(x) and s = u(t), then
(df)/(dx) = (d("wou"))/(dt) . (dt)/(dx) = (dw)/(ds) . (ds)/(dt) . (dt)/(dx)
provided all the derivatives in the statement exist. Reader is invited to formulate chain rule for composite of more functions.

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Derivative of composite function-chain rule [00:16:26]
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