How do you find the second derivative of #y^2=x^3#?

1 Answer
Jul 29, 2015

#y''= (3)/(4sqrtx)#

Explanation:

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.