RULE OF DIFFERENTIATION

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Vinegar purity

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Long Division

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Unit 1 - Mapping The Sociological Imagination

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levers

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ASSIGNMENT 2 - PRE FINAL BUSINESS MATHEMATICS II NAME : WAFIQAH BINTI WAHAB NO. MATRIC : 052870 PROGRAMME : DIPLOMA IN BANKING LECTURE NAME : MADAM HARDAYANNA ABD RAHMAN DUE DATE : THURSDAY, 9 JULY 2020

RULE OF DIFFERENTIATION

Main topic

RULE 5 : CHAIN RULE

= 15 (3x - 3)4

= 5 (3x - 3)4 (3)

Solution : f' (x) = 5 (3x-3)4 (3x-3)1

Example : f (x) = (3x - 3)5

f' (x) = n (ax + b)n-1 (ax + b)1

f (x) = (ax + b)n

RULE 6 : PRODUCT RULE

= -24x2 + 20x + 12

= -16x2 + 20x - 8x2 + 12

f' (x) = (4x - 5)(-4x) + (-2x2 + 3)(4)

v' = -4x

v = -2x2 + 3

u' = 4

Solution : u = 4x - 5

Example : (4x - 5) (-2x2 + 3)

f' (x) = uv' + vu'

h(x) = u, g(x) = v

f (x) = h(x)g(x)

RULE 7 : QUOTIENT RULE

= 3 / (2x +1)2

= 6x + 3 - 6x / (2x + 1)2

f' (x) = (2x + 1)3 - 3x(2) / (2x + 1)2

v' = 2

v = 2x + 1

u' = 3

Solution : u = 3x

Example : 3x / 2x + 1

f'(x) = vu' - uv' / v2

h(x) = u, g(x)

f (x) = h(x) / g(x)

RULE 4 : SUM RULE

Solution : f' (x) = 4x3 + 4

Example : f (x) = x4 + 4x

f (x) = h' (x) + g' (x)

f (x) = h (x) + g (x)

RULE 3 : POWER RULE

Solution : 9x8

Example : x9

f' (x) = nxn-1

f (x) = xn

Solution : f' (x) = 8

Example : f (x) = 8x

f' (x) = m

f (x) = mx

RULE 1 : CONSTANT RULE

Solution : f' (x) = 0

Example : f (x) = 13

f' (x) = 0

y = f (x) = c