You can use implicit differentiation:
#D(y^2)=D(x^3)#
#2yy'=3x^2#
#y'=(3x^(2))/(2y)#
Use the quotient rule:
#y''=[2y.6x-3x^(2).2y']/(4y^(2))#
Subs for #y'rArr#
#y''=[12xy-6x^(2)(3x^(2)/(2y))]/(4y^2)#
#y''=[12xy-9x^(4)/y]/(4y^2)#
#y''=(12xy^(2))/(4y^(3))-(9x^4)/(4y^(3))#
#y^(2)=x^(3)rArr#
#y''=(12x^(4))/(4y^(3))-(9x^(4))/(4y^(3))#
#y''=(3x^(4))/(4y^(3))#
Since #y^(2)=x^(3)# this becomes:
#y''=(3x^(4))/(4yx^(3))#
#y''=(3x)/(4y)#
#y^2=x^3#
#y=x^(3/2)#
Substituting for y gives:
#y''=(3x)/(4x^(3/2))#
#y''=(3)/(4sqrtx)#
You can also use explicit differentiation:
#y^(2)=x^(3)#
#y=sqrt(x^(3))#
#y=x^(3/2)#
Apply the power rule:
#y'=3/2.x^(1/2)#
And again for the 2nd derivative:
#y''=3/4x^(-1/2)#
#y''=3/4.(1)/(x^(1/2)#
#y''=(3)/(4sqrtx)#
Either way gets you there.
