# How do you find ((d^2)y)/(dx^2)?

## Question: Find (d^y)/(dx^2) by implicit differentiation when 2xy+2y^2=13

##### 2 Answers
Nov 3, 2016

$y ' ' = \pm \frac{13}{26 + {x}^{2}} ^ \left(\frac{3}{2}\right)$

#### Explanation:

Defining

$f \left(x , y \left(x\right)\right) = 2 x y \left(x\right) + 2 y {\left(x\right)}^{2} - 13 = 0$ then

$\frac{\mathrm{df}}{\mathrm{dx}} = 2 y + 2 x y ' + 4 y y ' = 0$ and

$\frac{d}{\mathrm{dx}} \frac{\mathrm{df}}{\mathrm{dx}} = 4 y ' + 2 x y ' ' + 4 y {'}^{2} + 4 y y ' ' = 0$

after substitution of $y '$ we have

$y ' ' = \frac{2 x y + 2 {y}^{2}}{x + 2 y} ^ 3 = \frac{13}{x + 2 y} ^ 3$

but $y = \frac{1}{2} \left(- x \pm \sqrt{26 + {x}^{2}}\right)$ so finally

$y ' ' = \pm \frac{13}{26 + {x}^{2}} ^ \left(\frac{3}{2}\right)$

Nov 3, 2016

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{13}{x + 2 y} ^ 3$

#### Explanation:

$2 x y + 2 {y}^{2} = 13$

Differentiating wrt $x$ and applying the product rule gives us:

$2 \left\{\left(x\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + \left(1\right) \left(y\right)\right\} + 4 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$x \frac{\mathrm{dy}}{\mathrm{dx}} + y + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0 \implies \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y}{x + 2 y}$

Differentiating again wrt $x$ and applying the product rule (twice) gives us:

$\therefore \left\{\left(x\right) \left(\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}\right) + \left(1\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\right\} + \frac{\mathrm{dy}}{\mathrm{dx}} + 2 \left\{\left(y\right) \left(\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}\right) + \left(2 \frac{\mathrm{dy}}{\mathrm{dx}}\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\right\} = 0$

$\therefore x \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{\mathrm{dy}}{\mathrm{dx}} + 2 y \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + 2 {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = 0$
$\therefore x \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + 2 \frac{\mathrm{dy}}{\mathrm{dx}} + 2 y \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + 2 {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = 0$

$\therefore \left(x + 2 y\right) \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + 2 \left(- \frac{y}{x + 2 y}\right) + 2 {\left(- \frac{y}{x + 2 y}\right)}^{2} = 0$
$\therefore \left(x + 2 y\right) \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{2 y}{x + 2 y} - \frac{2 {y}^{2}}{x + 2 y} ^ 2$
$\therefore \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{2 y}{x + 2 y} ^ 2 - \frac{2 {y}^{2}}{x + 2 y} ^ 3$

Putting over a common denominator gives:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{\left(2 y\right) \left(x + 2 y\right) - \left(2 {y}^{2}\right)}{x + 2 y} ^ 3$
$\therefore \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{\left(2 x y + 4 {y}^{2} - 2 {y}^{2}\right)}{x + 2 y} ^ 3$
$\therefore \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{\left(2 x y + 2 {y}^{2}\right)}{x + 2 y} ^ 3$
$\therefore \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{13}{x + 2 y} ^ 3$ (as $2 x y + 2 {y}^{2} = 13$)