You use the chain rule when functions are concatenated. This means that one function encloses a second function. For example f(x)=u(v(x)). The function v(x) is inside the function u(x).
Functions
A reminder on function notation will be useful.
A function is simply a set of instructions that are applied to the term inside the function. So a function given as u(x) tells you what to do to x. If instead you have the same function given as u(y), the instructions are applied to y.
Example 1
A function is given as u(x)=3x2+4x−2. Find an expression for u(y).
All you do here is replace each instance of x with an instance of y. Think of the term inside the function u as a placeholder. Hence
u(y)=3y2+4y−2
Example 2
A function is given as u(x)=3x2+4x−2. Another function is given as v(x)=2x−1. Find an expression for u(v(x)).
Recall that the term inside the definition of a function is just a placeholder. So
u(v(x))=3v(x)2+4v(x)−2
But you know that v(x)=2x−1, so this expression above can be written as
u(v(x))=3(2x−1)2+4(2x−1)−2
This is expanded and simplified to become
u(v(x))=12x2−4x−3
It should be noted that you would not be expected to spot the functions u(x) and v(x) from this result.
The chain rule
The chain rule concerns itself with differentiation functions of the form f(x)=u(v(x)), in other words a chain of functions. In particular, unlike with example 2, the concatenation may not be easily expanded and simplified.
PROCEDURE
1.
Identify the outer functionu(x) and inner functionv(x)individually.
2.
Using differentiation, find u′(x) and v′(x).
3.
The derivative of f(x) is defined as f′(x)=u′(v(x))v′(x). In words this is the derivative of u with respect to x evaluated at v(x), multiplied by the derivative of v with respect to x.
Note: There are two ways to think of u′(v(x)). One way is described in the procedure above: it's the derivative of u with respect to x, but evaluated at v(x). Alternatively it is the derivative of u with respect to v(x).
Alternative notation
Rather than using function notation when differentiating, you can use dxdy notation. For y=f(u) where u=g(x), the chain rule can be given as
dxdy=dudy×dxdu
This can be viewed as a substitution: you substitute some function u into your equation before differentiating.
Example 3
Differentiate y=(4x5+3x2+1)7 with respect to x.
Make the substitution u=4x5+3x2+1 and rewrite the equation: y=u7. Now find dudy and dxdu:
The inner function v(x) is x2−3. The outer function is u(x)=x3. Hence u′(x)=3x2 and v′(x)=2x. Thus
f′(x)=u′(v(x))3⋅(x2−3)2⋅v′(x)2x
Simplify:
f′(x)=6x(x2−3)2
Example 5
Find the first derivative of the functionf(x)=4x−31.
Identify the inner and outer functions: inner: v(x)=4x−3; outer: u(x)=x1=x−1. Thus the derivatives of these are v′(x)=4 and u′(x)=−x−2=−x21. Thus, using the chain rule f′(x)=u′(v(x))v′(x) gives
f′(x)=−v(x)21⋅4=−(4x−3)24
If necessary, this could be expanded and simplified further.