9.4 First Derivatives of the Quotient of Two Polynomials

9.4 Find the Derivatives of a Quotient using Quotient Rule

Method 1
The Quotient Rule

Example:


Method 2 (Differentiate Directly)



Example:
Given that  y = x 2 2 x + 1 ,  find  d y d x
Solution:
y = x 2 2 x + 1 d y d x = ( 2 x + 1 ) ( 2 x ) x 2 ( 2 ) ( 2 x + 1 ) 2      = 4 x 2 + 2 x 2 x 2 ( 2 x + 1 ) 2 = 2 x 2 + 2 x ( 2 x + 1 ) 2


Practice 1:
Given that  y = 4 x 3 ( 5 x + 1 ) 3 ,  find  d y d x
Solution: y = 4 x 3 ( 5 x + 1 ) 3 d y d x = ( 5 x + 1 ) 3 ( 12 x 2 ) 4 x 3 .3 ( 5 x + 1 ) 2 .5 [ ( 5 x + 1 ) 3 ] 2      = ( 5 x + 1 ) 3 ( 12 x 2 ) 60 x 3 ( 5 x + 1 ) 2 ( 5 x + 1 ) 6      = ( 12 x 2 ) ( 5 x + 1 ) 2 [ ( 5 x + 1 ) 5 x ] ( 5 x + 1 ) 6      = ( 12 x 2 ) ( 5 x + 1 ) 2 ( 1 ) ( 5 x + 1 ) 6      = 12 x 2 ( 5 x + 1 ) 4

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