Question

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

Answer 1

`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.