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
Calculus Topics


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