Maharashtra State BoardHSC Arts 12th Board Exam

Derivatives of Composite Functions - Chain Rule


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: (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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