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

2 Answers
Jan 21, 2016

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

Explanation:

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

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

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

Jan 21, 2016

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

Explanation:

#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#