The * Chain Rule* computes the derivative of the
* composite* of two functions. The * composite* is just "g inside
f" --- that is,

(Note that this is *not* multiplication!)

Here are some examples:

Here's a more complicated example:

One way to tell which function is "inside" and which is "outside" is to think about how you would plug numbers in. For example, take . What would you do to compute on your calculator? First, you'd square 1.7 --- . Next, you'd take the sine of that --- .

The function you did first --- squaring --- is the *inner*
function. The function you did second --- sine --- is the
*outer* function.

* Example.* Suppose

Compute , , and .

* Theorem.* (* Chain Rule*)
If f is differentiable at a and g is differentiable at , then the composite function is differentiable at a, and its derivative
is

In functional form, this is

In words, you differentiate the outer function while holding the inner function fixed, then you differentiate the inner function.

The proof is pretty technical, and you can omit it if you're taking a typical first-term calculus course. It is given at the end. In the examples, I'll focus on how you use the Chain Rule to compute derivatives.

* Example.* Compute .

looks like this:

Differentiate the outer function , obtaining . What is "junk"? It's . The first term in the Chain Rule is . (Notice that I differentiated the outer function, temporarily leaving the inner one untouched.)

Next, differentiate the inner function. The derivative of is .

Therefore,

* Example.* Compute .

While it would be correct to use the Quotient Rule, it's unnecessary.

That is,

In general, you do not need to use the Quotient Rule to differentiate things of the form

In the first case, use the Chain Rule as above. In the second case, divide the top by the number on the bottom.

* Example.* Compute .

In some of the examples which follow, I'll use the derivative formulas for and . They are:

* Example.* Compute .

* Example.* Compute .

Therefore,

* Example.* f and g are differentiable functions.
A table of some values for these functions is shown below.

Find .

By the Chain Rule,

* Example.* Compute .

* Example.* (a) Compute .

(b) Draw a picture to show the difference between the functions and , considered as composites of and .

(a)

(b) Here's a picture showing the difference between and :

In the first case, the outer function is the squaring function; in the second case, the outer function is the sine function.

* Example.* The derivative formulas for and are

Taking these for granted, find:

(a) .

(b) .

(a)

(b)

* Example.* Compute .

Differentiate from the outside in:

* Example.* Where does the graph of have a horizontal tangent?

Set and solve for x:

* The proof of the Chain Rule.*

This section is fairly technical, so you can probably skip it if you're reading this for first-term calculus.

* Lemma.* If f is differentiable at a, there is a
continuous function which satisfies:

(a) .

(b)

* Proof.* Define

Then

Thus,

Hence, p is a continuous function. This proves (a).

Note that for ,

For , this equation is true, since both sides are 0. This proves (b).

* Theorem.* (* Chain Rule*)
Suppose that . Assume f is differentiable at a and g is
differentiable at . Then the composite is differentiable at a, and

* Proof.* By the lemma, there are functions p and
q such that

Here . Thus, as , I have . By the rule for the limit of a composite, this means that as , I have .

The next few steps may be a little hard to follow, so I'll give some detail before I do the computation.

I will take the equation and substitute as follows:

1. On the left side, I'll plug in .

2. On the right side I'll plug in in for k.

Now here's the computation:

Now take the limit as on both sides. Remember that as , I have both and .

Copyright 2018 by Bruce Ikenaga