Suppose that f(x) and g(x) are functions that we know how to differentiate
and h(x) = f(g(x)). Since h(x) is defined as the composition of
f and g, it fol lows that the derivative of h(x) must somehow be related
to the derivatives of f and g. In this section we consider the problem
of differentiating functions defined as such (as the composition of two
functions).
1. The Statement of the Chain Rule
We start with an example.
Example 1.1. Find the derivative of the function
First note that h(x) is the composition of the functions g(x) = x+1 and
,
i.e. f(g(x)) = h(x). This means that the derivative of
h(x) must somehow be related to the derivatives of f(x) and g(x). Next
observe that the graph of h (x) is the same as that of f(x), except it has
been shifted over to the left by a distance of 1 unit. This means that the
derivative of h(x) must be the same as that of f(x), just shifted over
by 1 unit. Since
it follows that
Notice that this is very similar to the derivative of f(x).
In the previous example we de termined the derivative of a composition
of functions using geometric arguments. In general, we want a rule to
do this. The rule used to differentiate compositions is called the chain
rule. Deriving the chain rule is much more difficult that the other rules
we have considered, so we shall simply just state it and look at some
examples in detail.
Result 1.2. (The Chain Rule) If f(x) and g(x) are differentiable and
h(x) = f(g(x)), then
In words, we say “derivative of the outside function, composed with
the inside function multiplied by the derivative of the inside function”
where we consider f(x) to be the outside function and g(x) to be the
inside function.
Remark 1.3. Note that to apply the chain rule, we need to be good
at decomposing functions i.e. writing a function h(x) as a composition
of two other functions g and f.
This means to calculate the derivative of a composition of functions,
we need to take the following steps :
(i ) Decompose h(x) into a product of two functions f(x) and g(x)
i.e. h(x) = f(g(x)).
(ii ) Determine derivatives f′(x) and g′(x) and the composition
f′(g(x)).
(iii ) Apply the formula: h′(x) = f′(g(x))g′(x).
We illustrate with some examples.
Example 1.4. Differentiate the following functions.
We have
outside function:
inside function:
Therefore, applying the chain rule, we have
We have
outside function:
inside function:
Therefore, applying the chain rule, we have
We have
outside function:
inside function:
Therefore, applying the chain rule, we have
2. Applications of the Chain Rule
As well as differentiating basic compositions, the chain rule can be used
to determine derivatives of functions we cannot yet differentiate as well
as determine some new rules. We illustrate with a couple of specific
examples.
(i ) Differentiating an Exp onential Function of Base a
Suppose
.
From previous results, we know that
and so we can apply the chain rule to
differentiate f(x). Specifically, we have
outside function:
inside function:
Therefore, applying the chain rule, we have
(ii ) The General Exponential Rule
Suppose h(x) = ef(x). We can apply the chain
rule to determine
a rule which will allow us to differentiate this function
and leave the answer in terms of f(x) and its derivative.
Specifically, we have
outside function: 
inside function: 
Therefore, applying the chain rule, we have
3. Further Examples
We finish with some further examples of the chain rule.
Example 3.1. Determine a formula to differentiate
the composition
of three functions k(x) = f(g(h(x))).
We have
outside function: 
inside function: 
Notice that to calculate the derivative of the “inside”
function, we
needed to apply the chain rule. Now applying the chain rule, we have
Example 3.2. Differentiate
We have
outside function: 
inside function: 
Notice that to calculate the derivative of g(x) we needed
to apply the
chain rule. Specifically, we have
outside function: 
inside function: 
Now applying the chain rule, we have
Example 3.3. Some values of f (x), g(x), f′(x), and
g′(x) are given in
the table below.
Let 
Determine the following derivatives:
(i ) 
We have 
(ii ) 
We have 
(iii ) 
We have 
(iv) 
We have
We have 
Example 3.4. Suppose that the radius of a ball is
given as a function
of time and the graph is given below. Use this to determine V ′(1), the
rate of change of volume with respect to time when t = 2.
We know that V (r) = 
,
and we know that r is a function of t, so it
follows that V is a function of t too i.e. V (r(t)). Applying the chain
rule, it follows that 
. At t = 1,
looking at the graph, we have r = 4 and r′(t) = 2 (the slope at that
point). Thus we have V ′(2) = 16π.
