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 𝒚= 𝟏+𝟑𝒙 𝟒 𝟐 𝒙 𝟐 −𝟏 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.
𝒚= 𝟏+𝟑𝒙 𝟒 𝟐 𝒙 𝟐 −𝟏 𝟐 = 𝟏+𝟑(𝟎) 𝟒 𝟐 (𝟎) 𝟐 −𝟏 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 = 𝟐𝒙+𝟏 𝒙
𝒇 𝒙 = 𝟐𝒙+𝟏 𝒙 𝟏 𝟐
Rewrite
𝑓 ′ 𝑥 = times the parentheses to –1 times the derivative of what's in the parentheses (quotient rule will be needed)
𝒇 ′ 𝒙 = 𝟏 𝟐 𝟐𝒙+𝟏 𝒙 − 𝟏 𝟐 𝒙 𝟐 −(𝟐𝒙+𝟏)(𝟏) 𝒙 𝟐
𝒇 ′ 𝒙 = 𝟏 𝟐 𝟐𝒙+𝟏 𝒙 − 𝟏 𝟐 𝟐𝒙−𝟐𝒙−𝟏 𝒙 𝟐
𝒇 ′ 𝒙 = 𝟏 𝟐 𝒙 𝟐𝒙+𝟏 𝟏 𝟐 −𝟏 𝒙 𝟐
𝒇 ′ 𝒙 = −𝟏 𝟐 𝒙 𝟐 𝒙 𝟐𝒙+𝟏