# How do you find (d^2y)/(dx^2) for -2y^2+2=3x?

Nov 7, 2016

Please see the explanation section below.

#### Explanation:

$- 2 {y}^{2} + 2 = 3 x$

Differentiate w.r.t. $x$ to get

$- 4 y \frac{\mathrm{dy}}{\mathrm{dx}} = 3$

So, $\text{ }$ $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3}{4} {y}^{-} 1$

Differentiate again w.r.t. $x$ and we find

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{3}{4} {y}^{-} 2 \frac{\mathrm{dy}}{\mathrm{dx}}$

Replace $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3}{4} {y}^{-} 1$ and simplify

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{9}{16 {y}^{3}}$