CHAIN 1
THE RULE
The whole purpose of the chain rule is to be able to find the derivative of a complicated function without to much stress. Consider the following composition of functions: p (x) = f (g (k (h (x)))). The derivative p ' (x) = f ' (g (k (h (x))))* g ' (k (h (x)))* k ' (h (x))* h ' (x). Determining this derivative is like peeling an onion from the outside in. Take the derivative of the outer function, times the derivative of the next layer. Keep doing this until you get to the final inner-most layer. The Chain Rule is formally stated and proved at the end of these notes. EXAMPLE 1: SOLUTION:
Let u = x 4 2 x 2 - 6 and f u = u 100 . Then u ' = 4 x 3 4 x
Find f ' for f (x) = (x4 + 2x2 - 6)100.
f ' u = 100 u99 f ' x = f ' u x u ' x = 100 u 100 4 x 3 4 x = 100 x 4 2 x 2 - 699 4 x 3 4 x
Remember the rule for f (x) = xn for which f ' (x) = n xn 1? n can be any real number, even a fraction. Notice, that the final answer in Example 1 is the derivative of the inside times the derivative of the outside. Using the chain rule is like peeling an onion. Start with the outer-most function and take its derivative, then the next function, and so on. The next example illustrates the Chain Rule in combination with the product rule. It is very important that you gain facility in applying multiple rules in combination to the point that it becomes second nature to you! EXAMPLE 2:
1 -3 = x - 2 x 5 3 x - 2 x 5
Find f ' for f x =
SOLUTION:
Use the product rule to find u '. (f (x) = h (x) *g (x). f ' (x) = h ' (x)*g (x) + h (x)*g ' (x).)
2 EXAMPLE 3:
Find f ' for f x =
3 x 2 3 x 4.
SOLUTION: 1 Note that in this problem we have the function x 2 3 x 4 raised to the 3 power. This is thus a composite function which requires the Chain Rule. In differentiating the outer function we have to use the "drop down" rule for the outer function. We begin by letting u represent the the inner function. This is a procedure you should write out each time you do this type of problem, at least for a period of time when your "training wheels" are attached. Later you will be doing this type of problem in one step!
EXAMPLE 4: SOLUTION:
Find f ' for f x = sincos x 2 - 5 x 6.
This is a triple composite function. We begin by letting u represent the inner most function, x - 5 x 6 . Moving outward one function, we let p represent the cosine function. Finally we have the function f as y = f p = sin p. Thus we will be writing the derivative as
2
dy dy dp du = dx dp du dx
Using the Newton notation, this is equivalent to
f ' x = f ' pp ' u u ' x
3 EXAMPLE 5: SOLUTION: This is another triple composition like the one in Example 4. You might try this one on your own before peeking at the solution below.
Find f ' x for f x = tan x 4 x 3 .
2
You might want to simplify the above result, but it is not necessary. In a given problem situation in the future, the amount and nature of derivative simplifications depends upon the purpose to which the derivative is being applied. EXAMPLE 6: SOLUTION: Find
f ' for f x = 1 cos 2 x- 4 .
EXAMPLE 7: SOLUTION:
Find f ' x for f x = csc x cot x - 1 .
4 We end this series of examples with another triple composite function. EXAMPLE 8: SOLUTION:
Find f ' for f x = sincos 2 x - 5 .
Work through the examples provided in this set of supplemental notes. While doing this, keep in mind the fact that you will start with the derivative of the outside function, and keep taking the derivative of each of the following layers. It is like a multi-layered onion. Once you peeled the outer skin off, there is another skin.
Statement and Proof of the Chain Rule
Theorem: (Chain Rule)
Let y = f (x ) be a differentiable function of x and x = g(t ) be a differentiable function of t. Then y = F( t ) = f ( g ( t ) ) is also a differentiable function, and
F ' t = f ' g t g ' t or
dy dy dx = dt dx dt
Note: The Chain Rule statements above are equivalent with the one first using Newton's notation for the derivative and the second statement using Leibnitz' notation for the derivative. The "chaining" together of the two fractional Leibnitz' notation for the derivatives, "linked" together by the common dx is the motivation for the name "Chain Rule".
Proof:
Define =
lim
y - f ' x Since f is a differentiable function, by the definition of derivative, x
y = f ' x so it is clear that 0 as x 0. Solving for y we see that x 0 x y = f ' x x + x . Now we divide through this equation by t to obtain
y = t
f ' x
x x + t t
(1)
We are just about to let t approach 0, but we must first establish what happens to x . Consider the following.
t 0
lim x = lim
x 0
x t t
5 Since x = g( t ) is a differentiable function of t, the first factor in the limit exists and is g' ( t ). Thus we have
t 0
lim x = g ' t 0 = 0
Referring back to equation (1) we see that
lim
t 0
y = t
f ' x lim
t 0
x t
+
t 0
lim lim
t 0
x t
(2)
x y dx = g ' t = and lim = 0, it is clear that lim exists. Therefore dt t 0 t t 0 t 0 t equation (2) results in
Since
lim
dy dx dx = f ' x + 0 dt dt dt dy dy dx = dt dx dt
dy dx with F ' t , x with g t , and with g ' t we get the following expression for dt dt the Chain Rule:
If we replace
F ' t = D t f gt = f ' gt g' t
Extending the Chain Rule
The Chain Rule tells us how to determine the derivative of a composition of two functions. It is easily extended to multiple compositions of functions. Consider the function
y = F x = f g h x = f g hx
where f, g, and h are all differentiable functions. We can let y = f (u ), u = g (v ), and v = h (x ). This can be considered to be a chain of linked functions. Using the Leibnitz notation, we have
F ' x =
expressing the fact that we have
dy dy du dv = dx du dv dx
dy is the product of three derivatives. Using the Newton notation instead, dx dy du dv = f ' u , = g ' v , and = h ' x so that du dv dx
F ' x = f ' u g ' v h ' x
Writing everything in terms of x we have
D x f g h x = f ' g h x g ' h x h ' x
Example 4 above was done using the extended Chain Rule for a triple composition of functions.
6 Two final examples should cement your understanding of the Chain Rule. EXAMPLE 9: SOLUTION: This is another triple composition. We will list them before applying the Chain Rule.
y = f u = u ,
3
Find the derivative of y = sin 3 2 x 4 x 2 .
u = g v = sin v ,
v = hx = 2 x x
4
2
Thus f is the cubing function, u is the sine function, and v is the polynomial function in x. f is the outermost function; g is the middle function; h is the innermost function. We are to determine the derivative of this triple composition.
y = f g h x
The best way to apply the Chain Rule in practice is shown below since it is most easily remembered by students in the long run. dy dy du dv = dx du dv dx
dy 2 = 3u , du multiply the derivatives.
Then we have
du = cos v , dv
dv 3 = 8 x 2 x . Applying the Chain Rule we next dx
dy = 3 u 2 cos v 8 x 3 2 x dx
Finally we write everything strictly in terms of x.
dy 2 3 = 3 sin v cos v 8 x 2 x dx
= 3sin 2 x x cos 2 x x 8 x 2 x
2 4 2 4 2 3
. EXAMPLE 10: SOLUTION:
Find the derivative of y = sin 3 2 x 4 x 3
Here we have a quadruple composite function! With enough practice you should be able to do these derivatives "on the fly" without having to identify with a letter each individual function. You
7 start differentiating the outermost function and then work your way inward, multiplying by the derivative of the next inner function. There will be four factors since there are four functions: first the cubing function, then the sine function, followed by the square root function with the innermost function the polynomial 2 x 4 x 3 .
2 dy 1 4 3 4 3 4 3 2 3 2 = 3 sin 2 x x cos 2 x x 2 x x 8 x 3 x dx 2
-1