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 thaty=x22x+1,finddydx 

Solution:

y=x22x+1dydx=(2x+1)(2x)x2(2)(2x+1)2=4x2+2x2x2(2x+1)2=2x2+2x(2x+1)2 

 

Practice 1:

 Given thaty=4x3(5x+1)3,finddydx


Solution:

 y=4x3(5x+1)3dydx=(5x+1)3(12x2)4x3.3(5x+1)2.5[(5x+1)3]2=(5x+1)3(12x2)60x3(5x+1)2(5x+1)6=(12x2)(5x+1)2[(5x+1)5x](5x+1)6=(12x2)(5x+1)2(1)(5x+1)6=12x2(5x+1)4

 

 

 

 

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=x22x+1, find dydx

Solution:

y=x22x+1dydx=(2x+1)(2x)x2(2)(2x+1)2 =4x2+2x2x2(2x+1)2=2x2+2x(2x+1)2



Practice 1:
Given that y=4x3(5x+1)3, find dydx

Solution:
y=4x3(5x+1)3dydx=(5x+1)3(12x2)4x3.3(5x+1)2.5[(5x+1)3]2 =(5x+1)3(12x2)60x3(5x+1)2(5x+1)6 =(12x2)(5x+1)2[(5x+1)5x](5x+1)6 =(12x2)(5x+1)2(1)(5x+1)6 =12x2(5x+1)4