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

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

U2 L8 Chain and Quotient Rule Factor out common factors 4 , (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

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

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 = 1 2 times the parentheses to 1 2 –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