Understanding the Chain Rule

Most students are first introduced to the chain rule when shown how to differentiate a function such as y = (3x – 2)5. The problem is that is tempting to try and fit all chain rule differentiations into that format, for example trying to differentiate e3x – 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 yu 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) = (3x – 2)5, and I substitute
x =2. First I calculate 3 × 2 – 2 = 4, then I calculate 45 = 1024. In other words, I have first substituted into the function 3x – 2, and then the result into the function x5. Using function notation, if f(x) = 3x – 2 and g(x)x5, then
(3x – 2)5g(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 = (3x – 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(x2), start with ux2

y = (tanx)4, start with u = tanx

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

ye3x + 4, rewrite as ye(3x + 4), and start with u = 3x + 4

y = {3 \over {{x^2} - 3}} , rewrite as y = {3 {({x^2} - 3)^{-1}}} , and start with ux2 – 3


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

{{dy} \over {dx}} = 2x\sin ({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}

 

 

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