Chain Rule

This discussion will focus on the Chain Rule of Differentiation. The chain rule allows the differentiation of composite functions, notated by fg . For example take the composite function (x + 3)2. The inner function is g = x + 3. If x + 3 = u then the outer function becomes f = u2.

This rule states that:

The derivative of a composite function is the derivative of the outer function multiplied by the derivative of the inner function


CHAIN RULE OF DIFFERENTIATION:


( fg )( x )= f ( g( x ) )·g( x )



Let's work some examples

To work these examples requires the use of various differentiation rules. If you are not familiar with a rule go to the associated topic for a review.


(5x + 3)2

Step 1: Identify the inner function and rewrite the outer function replacing the inner function by the variable u.

g = 5x + 3 Inner Function


u = 5x + 3 Set Inner Function to u


f = u2        Outer Function

Step 2: Take the derivative of both functions.

Derivative of f = u2


d dx u 2      Original


2 u 1 0   Power & Constant


2u


______________________________


Derivative of g = x + 3


d dx 5x+3 Original


d dx 5x+ d dx 3 Sum Rule


5 d dx x+3  Constant Multiple


5x0 + 0      Power & Constant


5

Step 3: Substitute the derivatives and the original expression for the variable u into the Chain Rule and simplify.


( fg )( x )= f ( g( x ) )·g( x )

2u(5)       Chain Rule


2(5x + 3)(5) Substitute for u


50x + 30   Simplify

If the expression is simplified first, the chain rule is not needed.

Step 1: Simplify

(5x + 3)2 = (5x + 3)(5x + 3)


25x2 + 15x + 15x + 9


25x2 + 30x + 9

Step 2: Differentiate without the chain rule.

d dx ( 25 x 2 +30x+9 ) Original


d dx 25 x 2 + d dx 30x+ d dx 9 Sum Rule


25 d dx x 2 +30 d dx x+ d dx 9 Constant Multiple


25(2x1) + 30x0 + 0    Power & Constant


50x + 30

Example 1:       1 x 2 x+14

Step 1: Identify the inner function and rewrite the outer function replacing the inner function by the variable u.

g = x2 - x + 14 Inner Function


u = x2 - x + 14 Set IF to u


f= u 1 2     Outer Function

Step 2: Take the derivative of both functions.

Derivative of f= u 1 2


d dx u 1 2 Original


1 2 u 3 2 Power


1 2 u 3 2


__________________________


Derivative of g = x2 - x + 14


d dx ( x 2 x+14 ) Original


d dx x 2 d dx x+ d dx 14  Sum/Diff Rule


2x1 - 1x0 + 0 Power/Constant


2x1

Step 3: Substitute the derivatives and the original expression for the variable u into the Chain Rule and simplify.


( fg )( x )= f ( g( x ) )·g( x )

1 2 u 3 2  ·2x1 Chain Rule


[ 1 2 ( x 2 x+14 ) 3 2 ]·( 2x1 ) Sub


[ 1 2 ( x 2 x+14 ) 3 2 ]·( 2x1 )

Example 2:       ( 12 x 3 x ) 2

Step 1: Identify the inner function and rewrite the outer function replacing the inner function by the variable u.

g=( 12 x 3 x ) Inner Function


u=( 12 x 3 x ) Set IF to u


f = u2      Outer Function

Step 2: Take the derivative of both functions.

Derivative of f = u2


d dx u 2     Original


2u1        Power


2u


________________________


Derivative of g= 12 x 3 x


d dx ( 12 x 3 x ) Original


Apply Chain Rule:


[ ( x ) d dx ( 12 x 3 ) ][ ( 12 x 3 ) d dx ( x ) ] x 2


[ ( x )( 6 x 2 ) ][ ( 12 x 3 )( 1 ) ] x 2  Take deriv.


[ 6 x 3 ][ 12 x 3 ] x 2   Simplify


4 x 3 1 x 2

Step 3: Substitute the derivatives and the original expression for the variable u into the Chain Rule and simplify.


( fg )( x )= f ( g( x ) )·g( x )

2u ·  4 x 3 1 x 2   Chain Rule


2( 12 x 3 x ) ·  4 x 3 1 x 2 Sub for u


( 24 x 3 )( 4 x 3 1 ) x 3





Related Links:
Math
algebra
Trigonometry Differentiation Rules
Inverse Trigonometric Differentiation Rules
Algebra Topics


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