The Chain Rule

The Chain Rule

Our goal is to differentiate functions such as

y = (3x + 1)10

 The Chain Rule  If            y = y(u)  is a function of  u, and            u = u(x)  is a function of x then            dy           dy        du                     =                                                   dx            du       dx

In our example we have

y  =  u10

and

u  =  3x + 1

so that

dy/dx  =  (dy/du)(du/dx)

=  (10u9) (3)  =  30u9  =  30 (3x+1)9

Proof of the Chain Rule

Recall an alternate definition of the derivative:

Examples

Find f '(x) if

1. f(x) = (x3 - x + 1)20

2. f(x) = (x4 - 3x3 + x)5

3. f(x) = (1 - x)9 (1-x2)4

4.                (x3 + 4x - 3)7
f(x)  =
(2x - 1)3

Solution:

1. Here

f(u) = u20

and

u(x) = x3 - x + 1

So that the derivative is

[20u19] [3x2 - 1]  =  [20(x3 - x + 1)19] [3x2 - 1]

2. Here

f(u) = u5

and

u(x) = x4 - 3x3 + x

So that the derivative is

[5u4] [4x3 - 9x2 + 1]  =  [5(x4 - 3x3 + x)4] [4x3 - 9x2 + 1]

3. Here we need both the product and the chain rule.

f'(x) = [(1 - x)9] [(1 - x2)4]' + [(1 - x)9] '  [(1 - x2)4]

We first compute

[(1 - x2)4] ' = [4(1 - x2)3] [-2x]

and

[(1 - x)9] '  = [9(1 - x)8] [-1]

Putting this all together gives

f'(x) = [(1 - x)9] [4(1 - x2)3] [-2x]  -  [9(1 - x)8]  [(1 - x2)4]

4. Here we need both the quotient and the chain rule.

(2x - 1)3 [(x3 + 4x - 3)7] '  -  (x3 + 4x - 3)7 [(2x - 1)3] '
f '(x) =
(2x - 1)6

We first compute

[(x3 + 4x - 3)7] ' = [7(x3 + 4x - 3)6] [3x2 + 4]

and

[(2x - 1)3] '  = [3(2x - 1)2] [2]

Putting this all together gives

7(2x - 1)3 (x3 + 4x - 3)6 (3x2 + 4)  +  6(x3 + 4x - 3)7 (2x - 1)2
f '(x) =
(2x - 1)6

Exercise

Find the derivative of

x2(5 - x3)4
f(x)  =
3 - x

Application

Suppose that you put \$1000 into a bank at an interest rate r compounded monthly for 3 years.  Then the amount A that will be in the account at the end of the three years will be

A = 1000(1 + r/12)36

Find the rate at which A rises with respect to a rise in the interest rate when the interest rate is 6%.

Solution

We are asked to find a derivative.  We use the chain rule with

u  =  1 + r/12        and        A(u)  =  1000u36

The two derivatives are

u'  =  1/12        and        A'  =  36000u35

The chain rule gives

dA/dr  =  dA/du du/dr  =  (1/12) 36000 u35

=  3000 u35  =  3000(1 + r/12)35

Now plug in   r   =   6%  =  0.06   to get

3000(1 + 0.06/12)35  =  3572.18