The Chain Rule

The Chain Rule

Our goal is to differentiate functions such as

y = (3x + 1)10

The last thing that we would want to do is FOIL this out ten times.  We now look for a better way.

 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           dy     du
=
dx           du     dx

= (10u9)(3) = 30(3x+1)9

Proof of the Chain Rule

Recall an alternate definition of the derivative: Example:

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

Solutions

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.  First the product rule

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

Now 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]

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

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