# How do you find (d^2y)/(dx^2) given y^2=(x-1)/(x+1)?

Dec 1, 2016

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

Alternatively:

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{2 x - 1}{{\left(x + 1\right)}^{4} {\left(\frac{x - 1}{x + 1}\right)}^{\frac{3}{2}}}$

#### Explanation:

${y}^{2} = \frac{x - 1}{x + 1}$

Differentiating the LHS implicitly and the RHS using the quotient rule wrt $x$ we have:

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

$\left(\begin{matrix}\text{NB If we wanted an explicit expression for "dy/dx" we have" \\ dy/dx = 1 / ((x+1)^2sqrt((x-1)/(x+1))) " as } y = \sqrt{\frac{x - 1}{x + 1}}\end{matrix}\right)$

We have established:

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

Differentiating the LHS of [1] Implicitly with the product rule, and the RHS using the chain rule we get:

$\left(y\right) \left(\frac{d}{\mathrm{dx}} \frac{\mathrm{dy}}{\mathrm{dx}}\right) + \left(\frac{d}{\mathrm{dx}} y\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = - 2 {\left(x + 1\right)}^{-} 3 \left(1\right)$
$\therefore y \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 + \frac{\mathrm{dy}}{\mathrm{dx}} \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2}{x + 1} ^ 3$
$\therefore y \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = - \frac{2}{x + 1} ^ 3$
$\therefore y \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 + {\left(\frac{1}{{\left(x + 1\right)}^{2} \sqrt{\frac{x - 1}{x + 1}}}\right)}^{2} = - \frac{2}{x + 1} ^ 3$
$\therefore y \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 + \frac{1}{{\left(x + 1\right)}^{4} \left(\frac{x - 1}{x + 1}\right)} = - \frac{2}{x + 1} ^ 3$

We can rearrange, and cancel a factor of $\left(x + 1\right)$:
$- y \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{1}{{\left(x + 1\right)}^{3} \left(x - 1\right)} + \frac{2}{x + 1} ^ 3$
$\therefore - y \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{1 + 2 \left(x - 1\right)}{{\left(x + 1\right)}^{3} \left(x - 1\right)}$
$\therefore - y \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{1 + 2 x - 2}{{\left(x + 1\right)}^{3} \left(x - 1\right)}$
$\therefore - y \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{2 x - 1}{{\left(x + 1\right)}^{3} \left(x - 1\right)}$
$\therefore y \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{2 x - 1}{{\left(x + 1\right)}^{3} \left(x - 1\right)}$

And, as before to get an explicit expression we divide by $y$ to get
$\therefore \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{2 x - 1}{{\left(x + 1\right)}^{3} \left(x - 1\right)} \cdot \frac{1}{\sqrt{\frac{x - 1}{x + 1}}}$

We can also write this as;

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{2 x - 1}{{\left(x + 1\right)}^{4} \frac{x - 1}{x + 1}} \cdot \frac{1}{\frac{x - 1}{x + 1}} ^ \left(\frac{1}{2}\right)$
$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{2 x - 1}{{\left(x + 1\right)}^{4} {\left(\frac{x - 1}{x + 1}\right)}^{\frac{3}{2}}}$