Presentation is loading. Please wait.

Presentation is loading. Please wait.

U2 L8 Chain and Quotient Rule CHAIN & QUOTIENT RULE

Similar presentations


Presentation on theme: "U2 L8 Chain and Quotient Rule CHAIN & QUOTIENT RULE"β€” Presentation transcript:

1 U2 L8 Chain and Quotient Rule CHAIN & QUOTIENT RULE
UNIT 2 LESSON 8 CHAIN & QUOTIENT RULE

2 U2 L8 Chain and Quotient Rule
Chain Rule and Quotient Rule Example 1 Find the derivative of π’š= 𝟏+πŸ‘π’™ πŸ’ 𝟐 𝒙 𝟐 βˆ’πŸ 𝟐 dy = (2x2 – 1)2 [4(1 + 3x)3(3)] – (1 + 3x)4[2(2x2 – 1)(4x)] dx [(2x2 – 1)2]2 dy = (2x2 – 1)2 [12(1 + 3x)3] – (1 + 3x)4[8x(2x2 – 1)] dx (2x2 – 1)4

3 U2 L8 Chain and Quotient Rule
Factor out common factors , (2x2 – 1), and (1 + 3x)3 dy = (2x2 – 1)2 [12(1 + 3x)3] – (1 + 3x)4[8x(2x2 – 1)] dx (2x2 – 1)4 dy = 4 (2x2 – 1) (1 + 3x)3[3(2x2 – 1) – 2x(1 + 3x)] dx (2x2 – 1)4 dy = 4 (2x2 – 1) (1 + 3x)3[6x2 – 3 – 2x – 6x2] dx (2x2 – 1)4 dy = 4 (2x2 – 1) (1 + 3x)3(– 3 – 2x) dx (2x2 – 1)4 dy = 4 (1 + 3x)3(– 3 – 2x) dx (2x2 – 1)3

4 U2 L8 Chain and Quotient Rule
Find the equation of the tangent line at x = 0. π’š= 𝟏+πŸ‘π’™ πŸ’ 𝟐 𝒙 𝟐 βˆ’πŸ 𝟐 = 𝟏+πŸ‘(𝟎) πŸ’ 𝟐 (𝟎) 𝟐 βˆ’πŸ 𝟐 =𝟏 Equation of tangent line 1 = 12(0) + b dy = 4 (1 + 3x)3(– 3 – 2x) dx (2x2 – 1)3 b = 1 dy = 4 (1 + 3(0))3(– 3 – 2(0)) dx (2(0)2 – 1)3 y = 12x + 1 Slope = dy = 4(1 )3(– 3) = 12 dx ( – 1)3

5 U2 L8 Chain and Quotient Rule
Chain Rule and the Quotient Rule Example 2 U2 L8 Chain and Quotient Rule 𝒇 𝒙 = πŸπ’™+𝟏 𝒙 𝒇 𝒙 = πŸπ’™+𝟏 𝒙 𝟏 𝟐 Rewrite 𝑓 β€² π‘₯ = times the parentheses to –1 times the derivative of what's in the parentheses (quotient rule will be needed) 𝒇 β€² 𝒙 = 𝟏 𝟐 πŸπ’™+𝟏 𝒙 βˆ’ 𝟏 𝟐 𝒙 𝟐 βˆ’(πŸπ’™+𝟏)(𝟏) 𝒙 𝟐 𝒇 β€² 𝒙 = 𝟏 𝟐 πŸπ’™+𝟏 𝒙 βˆ’ 𝟏 𝟐 πŸπ’™βˆ’πŸπ’™βˆ’πŸ 𝒙 𝟐 𝒇 β€² 𝒙 = 𝟏 𝟐 𝒙 πŸπ’™+𝟏 𝟏 𝟐 βˆ’πŸ 𝒙 𝟐 𝒇 β€² 𝒙 = βˆ’πŸ 𝟐 𝒙 𝟐 𝒙 πŸπ’™+𝟏


Download ppt "U2 L8 Chain and Quotient Rule CHAIN & QUOTIENT RULE"

Similar presentations


Ads by Google