What is the second derivative of #1/x^2#?

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Answer 1 · 2016-01-21T23:18:21

# f''(x) = 6/x^4 #

rewrite #f(x) = 1/x^2 = x^-2 #

# rArr f'(x) = -2x^-3 #

#rArr f''(x) = 6x^-4 = 6/x^4 #

Answer 2 · 2016-01-21T23:29:19

#y''=6x^-4=6/x^4#

#y=1/x^2#

#y''=d/(dx)y'=(d^2y)/dx^2#

To find #y'# e #y''# you can use the Power Rule:

given:

#y=x^n => y'=nx^(n-1)#

Remembering the power proprerties: #1/a^n=a^(-n)#

#y=1/x^2=x^(-2)=>y'=-2x^(-2-1)=-2x^-3#

#y''=d/(dx)(-2x^-3)=-2d/(dx)x^-3=-2*(-3)x^(-3-1)=#

#=6x^-4=6/x^4#