9.4 First Derivatives of the Quotient of Two Polynomials

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