How do you find the second derivative of y^2 - 16x = 0y216x=0?

1 Answer
Nov 22, 2016

(d^2y)/(dx^2)=(-64)/(y^3)d2ydx2=64y3

Explanation:

y^2-16x=0y216x=0

Implicitly differentiate to find first derivative:
2y(dy/dx)-16=02y(dydx)16=0
dy/dx(2y)=16dydx(2y)=16
dy/dx=16/(2y)dydx=162y
dy/dx=8/ydydx=8y

Implicitly differentiate to find second derivative (use quotient rule ):
(d^2y)/(dx^2)=frac{(y)(0)-(8)(dy/dx)}{y^2}d2ydx2=(y)(0)(8)(dydx)y2

Substitute in dy/dx=8/ydydx=8y
(d^2y)/(dx^2)=frac{-(8)(8/y)}{y^2}d2ydx2=(8)(8y)y2
(d^2y)/(dx^2)=frac{(-64)/y}{y^2}d2ydx2=64yy2
(d^2y)/(dx^2)=(-64)/(y^3)d2ydx2=64y3

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