What is the second derivative of #f(x)=(x^2-x^3)^(1/3)#?

1 Answer
Jan 30, 2016

Answer:

#1/3(2-6x)(x^2-x^3)^(-2/3){1-((4-6x))/((3x-3x^2))}#

Explanation:

here,
#f(x)=(x^2-x^3)^(1/3)#

so, the first derivative is,

#fprime(x)=d/(dx)(f(x))#

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

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

#=1/3(x^2-x^3)^(-2/3)(2x-3x^2)#

so the second derivative is,

#fprimeprime(x)=d/(dx)(fprime(x))#

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

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

#=1/3{(2x-3x^2)d/(dx)(x^2-x^3)^(-2/3)+(x^2-x^3)^(-2/3)d/(dx)(2x-3x^2)}#

# =1/3{-2/3(2x-3x^2)(x^2-x^3)^(-5/3)(2-6x)+(x^2-x^3)^(-2/3)(2-6x)} #

#=1/3(2-6x)(x^2-x^3)^(-2/3){-2/3((2x-3x^2)/(x^2-x^3))+1} #

#=1/3(2-6x)(x^2-x^3)^(-2/3){-2/3((cancel(x)(2-3x))/(cancel(x)(x-x^2))+1}#

#=1/3(2-6x)(x^2-x^3)^(-2/3){1-((4-6x))/((3x-3x^2))}#