With some functions and expressions, differentiation is a long process. e.g. Differentiating the function h(x) = (2x + 3)5 first needs a long expansion of the brackets.

If we think of h(x) = (2x + 3)5 as a combination of two functions with g(x) = 2x + 3 being inside the function f(x) so that
f(x) = (g(x))5. This is known as a function of a function or a composite function f o g.

To differentiate a function of a function we use the Chain Rule.

Chain Rule Formula

This states:

(f o g )'(x)   =   f '(g(x))  .  g'(x)
 

 

The Chain Rule looks complex, but with practice, which is essential, it should become more straightforward.

Example For our example, to differentiate h(x) = (2x + 3)5:

Let f(x) = x5 and g(x) = 2x + 3

Therefore, f '(x) = 5x4 and g '(x) = 2

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

5(2x + 3)4 . 2

= 10(2x + 3)4

Alternative Chain Rule formula

Another way to differentiate a composite function is to use Y12_The_Chain_Rule_01.gif notation, which behaves like fractions.

The Chain Rule states:

Y12_The_Chain_Rule_02.gif

Example For our example, to differentiate y = (2x + 3)5

Let u = 2x + 3, so y = u5
Therefore,Y12_The_Chain_Rule_03.gif = 2 and Y12_The_Chain_Rule_04.gif = 5u4
Using the chain rule Y12_The_Chain_Rule_05.gif
Y12_The_Chain_Rule_06.gif = 5u4 . 2
substituting u back in.
      = 10(2x + 3)4

Roots and Negative Indices using the Chain Rule

As with most differentiation, rational functions and expressions involving roots must be written in index form before differentiation.

Example 1

Differentiate Y12_The_Chain_Rule_07.gif

Write the function as y = (x2 + 3x)-1
Y12_The_Chain_Rule_08.gif

 

Example 2

Differentiate f(x) = √(3x2 + 2)

Write the expression as f(x) = Y12_The_Chain_Rule_09.gif
Y12_The_Chain_Rule_10.gif
            Y12_The_Chain_Rule_11.gif