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 2x+1 , find  dy dx
Solution:
y= x 2 2x+1 dy dx = (2x+1)(2x) x 2 (2) (2x+1) 2      = 4 x 2 +2x2 x 2 (2x+1) 2 = 2 x 2 +2x (2x+1) 2 

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