RULE OF DIFFERENTIATION

Mais informações

VM265 F13 Roberson Chara

por Chara Roberson

RULES OF DIFFERENTIATION

por azim farhan

RULES OF DIFFERENTIATION

por John Kelabu

Company wiki structure

por Kos Huang

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