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Most students are first introduced to the chain rule when shown how to differentiate a function such as *y* = (3*x* – 2)^{5}. The problem is that is tempting to try and fit all chain rule differentiations into that format, for example trying to differentiate *e*^{3x – 2} in the same way.

**What is the chain rule?**

It’s a calculus formula with a wide range of uses, just one of which is differentiating a ‘function of a function.’ Quite simply, differentiation concerns the rate at which one variable is changing compared to a second variable. The chain rule extends this so that we can calculate how three (or more) vary with each other.

Suppose three people, called Yasmin, Uther and Xavier, are running together. Yasmin runs twice as fast as Uther, and Uther runs three times as fast as Xavier. How fast is Yasmin running compared to Xavier? Not hard to see that this will be 2 × 3 = 6 times faster. Now, ${{dy} over {dx}}$ is the rate of change of *y* compared to *x*, and velocity is a rate of change. So if Yasmin, Uther and Xavier are reduced to the letters *y, u, *and *x*, we get:

${{dy} over {dx}} = {{dy} over {du}} times {{du} over {dx}}$

The chain rules simply states the multiplicative relationship between the rates of change of three quantities *y*, *u* and *x.*

**Function of a function**

Let’s leave the chain rule for a moment to understand the concept of ‘function of a function.’ Suppose I take the example of *f(x)* = (3*x* – 2)^{5}, and I substitute

*x* =2. First I calculate 3 × 2 – 2 = 4, then I calculate 4^{5} = 1024. In other words, I have first substituted into the function 3*x* – 2, and then the result into the function *x*^{5}. Using function notation, if *f(x)* = 3*x* – 2 and *g(x)* = *x*^{5}, then

(3*x* – 2)^{5} = *g(f(x)), *or in IB notation, (*g* º *f**)(x)*. Note that multiplication isn’t involved: in function of a function, the output of one function becomes the input of the next. I shall refer to these as the ‘inner’ and ‘outer’ functions.

**Using ****the**** chain rule to differentiate function of a function**

When faced with function of a function, we simplify things by replacing the inner function with a single letter, usually *u*. Now that we have three variables, we need the chain rule. Here’s the full working for differentiating *y* = (3*x* – 2)^{5}.

The procedure is the same every time you carry out a chain rule differentiation – the only thinking you really have to do is to identify the inner function. So, for:

*y* = sin(*x*^{2}), start with *u* = *x*^{2}

*y* = (tan*x*)^{4}, start with *u* = tan*x*

Sometimes the brackets are implied – just put them in to get started:

*y* = *e*^{3x + 4}, rewrite as *y* = *e*^{(3x + 4)}, and start with *u* = 3*x* + 4

$y = {3 over {{x^2} – 3}}$ , rewrite as $y = {3 {({x^2} – 3)^{-1}}}$ , and start with *u* = *x*^{2} – 3

If you’d like to try the four examples above, the differentiated functions are:

${{dy} over {dx}} = 2xsin ({x^2}) $

${{dy} over {dx}} = 4{sec ^2}x{(tan x)^3}{rm{ or }}4{sec ^2}x{tan ^3}x $

$ {{dy} over {dx}} = 3{e^{3x + 4}} $

${{dy} over {dx}} = {{ – 6x} over {{{({x^2} – 3)}^2}}}{rm{ or }} – 6x{({x^2} – 3)^{ – 2}} cr} $