Question

How do you find the second derivative of `y^2 - 16x = 0`?

Answer 1

`(d^2y)/(dx^2)=(-64)/(y^3)`

Explanation:

`y^2-16x=0`

Implicitly differentiate to find first derivative:
`2y(dy/dx)-16=0`
`dy/dx(2y)=16`
`dy/dx=16/(2y)`
`dy/dx=8/y`

Implicitly differentiate to find second derivative (use quotient rule ):
`(d^2y)/(dx^2)=frac{(y)(0)-(8)(dy/dx)}{y^2}`

Substitute in `dy/dx=8/y`
`(d^2y)/(dx^2)=frac{-(8)(8/y)}{y^2}`
`(d^2y)/(dx^2)=frac{(-64)/y}{y^2}`
`(d^2y)/(dx^2)=(-64)/(y^3)`

I am not sure if this answer can be simplified or not by substituting the value of y from the original equation...