Find the first four derivatives :

Question

i.) y = x^-3 + 3x^3 + 6x^2
ii.) y = - (3/x)
iii.) y = 2cos2x

Answer 1

i.) y = x^-3 + 3x^3 + 6x^2

y' = -3x^-4 + 9x^2 + 12x
y'' = 12x^-5 + 18x
y''' = -60x^-6
y'''' = 360x^-7

ii.) y = - (3/x)

y' = 3x^-2
y'' = -6x^-3
y''' = 18x^-4
y'''' = -72x^-5

iii.) y = 2cos2x

y' = -4sin2x
y'' = -8cos2x
y''' = 16sin2x
y'''' = 32cos2x