# U2 L8 Chain and Quotient Rule CHAIN & QUOTIENT RULE

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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 π= π+ππ π π π π βπ π dy = (2x2 β 1)2 [4(1 + 3x)3(3)] β (1 + 3x)4[2(2x2 β 1)(4x)] dx [(2x2 β 1)2]2 dy = (2x2 β 1)2 [12(1 + 3x)3] β (1 + 3x)4[8x(2x2 β 1)] dx (2x2 β 1)4

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

U2 L8 Chain and Quotient Rule
Find the equation of the tangent line at x = 0. π= π+ππ π π π π βπ π = π+π(π) π π (π) π βπ π =π Equation of tangent line 1 = 12(0) + b dy = 4 (1 + 3x)3(β 3 β 2x) dx (2x2 β 1)3 b = 1 dy = 4 (1 + 3(0))3(β 3 β 2(0)) dx (2(0)2 β 1)3 y = 12x + 1 Slope = dy = 4(1 )3(β 3) = 12 dx ( β 1)3

U2 L8 Chain and Quotient Rule
Chain Rule and the Quotient Rule Example 2 U2 L8 Chain and Quotient Rule π π = ππ+π π π π = ππ+π π π π Rewrite π β² π₯ = times the parentheses to β1 times the derivative of what's in the parentheses (quotient rule will be needed) π β² π = π π ππ+π π β π π π π β(ππ+π)(π) π π π β² π = π π ππ+π π β π π ππβππβπ π π π β² π = π π π ππ+π π π βπ π π π β² π = βπ π π π π ππ+π

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