# How do you find (d^2y)/(dx^2) for 2x-5y^2=3?

##### 2 Answers
Jul 21, 2017

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{1}{25 {y}^{3}}$

#### Explanation:

$2 x - 5 {y}^{2} = 3 \implies \frac{d \left(2 x\right)}{\mathrm{dx}} - \frac{d \left(5 {y}^{2}\right)}{\mathrm{dx}} = \frac{d \left(3\right)}{\mathrm{dx}} \implies$

$2 - 10 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0 \implies 1 - 5 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0 . . \left(1\right)$

$\frac{d \left(1\right)}{\mathrm{dx}} - \frac{d \left(5 y \frac{\mathrm{dy}}{\mathrm{dx}}\right)}{\mathrm{dx}} = 0 \implies - \frac{d \left(5 y \frac{\mathrm{dy}}{\mathrm{dx}}\right)}{\mathrm{dx}} = 0 \implies$

$\frac{d \left(5 y\right)}{\mathrm{dx}} \frac{\mathrm{dy}}{\mathrm{dx}} + 5 y \frac{d \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{\mathrm{dx}} = 0 \implies$

$5 {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} + 5 y \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 0$

Let's substitude from the equation $\left(1\right)$ that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{5 y}$

So we get :

$5 {\left(\frac{1}{5 y}\right)}^{2} + 5 y \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 0 \implies \frac{1}{5 {y}^{2}} + 5 y \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 0 \implies$

$5 y \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{1}{5 {y}^{2}} \implies \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{1}{25 {y}^{3}}$

Jul 21, 2017

${\left(\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}\right)}^{2} = \frac{1}{5} \frac{1}{2 x - 3} ^ 3$

#### Explanation:

Differentiate both sides of the equation with respect to $x$:

$\frac{d}{\mathrm{dx}} \left(2 x - 5 {y}^{2}\right) = 0$

$2 - 10 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$10 y \frac{\mathrm{dy}}{\mathrm{dx}} = 2$

$\left(1\right) \text{ } \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{5 y}$

and differentiating again:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{1}{5 {y}^{2}} \frac{\mathrm{dy}}{\mathrm{dx}}$

substitute now $\frac{\mathrm{dy}}{\mathrm{dx}}$ from $\left(1\right)$:

$\left(2\right) \text{ } \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{1}{25 {y}^{3}}$

We can also have an implicit equation for $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}$ by squaring both sides of $\left(2\right)$:

${\left(\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}\right)}^{2} = \frac{1}{{5}^{4} {y}^{6}} = \frac{1}{5} ^ 4 {\left(\frac{1}{y} ^ 2\right)}^{3}$

and substituting ${y}^{2} = \frac{2 x - 3}{5}$ from the original equation:

${\left(\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}\right)}^{2} = \frac{1}{5} \frac{1}{2 x - 3} ^ 3$

Function:

graph{y^2 = (2x-3)/5 [-10, 10, -5, 5]}

Derivative:

graph{y^2 = 1/5 1/(2x-3)^3 [-10, 10, -5, 5]}