# How do you find (d^2y)/(dx^2) for 2x+y^2=5?

Dec 4, 2016

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{1}{y} ^ 3$

#### Explanation:

If an equation does not express $y$ explicitly in terms of $x$ as in $y = f \left(x\right)$, and instead we have $g \left(y\right) = f \left(x\right)$, then when we differentiate we apply the chain rule so that we differentiate $g \left(y\right)$ wrt $y$ rather than $x$ as in:

$g \left(y\right) = f \left(x\right)$
$\therefore \frac{d}{\mathrm{dx}} g \left(y\right) = \frac{d}{\mathrm{dx}} f \left(x\right)$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} \frac{d}{\mathrm{dy}} g \left(y\right) = f ' \left(x\right)$
$\therefore g ' \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right)$

This will typically result in $\frac{\mathrm{dy}}{\mathrm{dx}}$ being a function of $x$ and $y$, rather than $\frac{\mathrm{dy}}{\mathrm{dx}}$ being a function of $x$ alone as we usually get

So for $2 x + {y}^{2} = 5$ we have:

$\frac{d}{\mathrm{dx}} \left(2 x\right) + \frac{d}{\mathrm{dx}} \left({y}^{2}\right) = \frac{d}{\mathrm{dx}} \left(5\right)$
 :. 2 + dy/dxd/dy(y^2) = 0)
$\therefore 2 + \frac{\mathrm{dy}}{\mathrm{dx}} 2 y = 0$
$\therefore 1 + \frac{\mathrm{dy}}{\mathrm{dx}} y = 0$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{y}$ ..... [1]

Differentiating again wrt $x$ we get:

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{d}{\mathrm{dx}} \left(- \frac{1}{y}\right)$
$\therefore \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{\mathrm{dy}}{\mathrm{dx}} \frac{d}{\mathrm{dy}} \left(\frac{1}{y}\right)$
$\therefore \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{\mathrm{dy}}{\mathrm{dx}} \left(- {y}^{-} 2\right)$
$\therefore \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{\mathrm{dy}}{\mathrm{dx}} \left(\frac{1}{y} ^ 2\right)$

And substituting from [1] we get:

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \left(- \frac{1}{y}\right) \left(\frac{1}{y} ^ 2\right)$
$\therefore \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{1}{y} ^ 3$