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

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

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Answer 1 · 2016-11-22T21:25:35

$\frac{d^{2} y}{{d x}^{2}} = \frac{- 64}{y^{3}}$ $(d^2y)/(dx^2)=(-64)/(y^3)$

Explanation:

$y^{2} - 16 x = 0$ $y^2-16x=0$

Implicitly differentiate to find first derivative:
$2 y (\frac{d y}{d x}) - 16 = 0$ $2y(dy/dx)-16=0$
$\frac{d y}{d x} (2 y) = 16$ $dy/dx(2y)=16$
$\frac{d y}{d x} = \frac{16}{2 y}$ $dy/dx=16/(2y)$
$\frac{d y}{d x} = \frac{8}{y}$ $dy/dx=8/y$

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

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

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

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

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Explanation:

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How do you find the second derivative of y^2 - 16x = 0y2−16x=0y^2 - 16x = 0?

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