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

Dec 22, 2016

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

#### Explanation:

It would be possible to explicit $y \left(x\right) = \pm \sqrt{\frac{2 x - 5}{5}}$, but we will use the procedure for implicit differentiation.

Differentiate the equation:

$2 x - 5 {y}^{2} = 5$

term by term with respect to $x$, considering that based on the chain rule:

$\frac{\mathrm{dg} \left(y\right)}{\mathrm{dx}} = \frac{\mathrm{dg} \left(y\right)}{\mathrm{dy}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

So, we have:

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

Solving for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{5 y}$

Differentiating again:

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

Now from the original equation we have:

$5 {y}^{2} = 5 - 2 x$

and we can use this to substitute $y$ in the expression of the second derivative:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{1}{25 {y}^{3}} = - \frac{1}{\sqrt{5}} \cdot \frac{1}{5 {y}^{2}} ^ \left(\frac{3}{2}\right) = - \frac{1}{\sqrt{5} {\left(5 - 2 x\right)}^{\frac{3}{2}}}$