# Given that  x^2 - y^2 =25  then show (d^2y)/(dx^2) =-25/y^3 ?

Sep 25, 2017

First a point on notation , we solve equations but we derive (or show) expressions.

So we are asked to show that given:

${x}^{2} - {y}^{2} = 25$ ..... [A]

Then

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = f ' ' \left(x , y\right) = - \frac{25}{y} ^ 3$

We proceed by differentiating the initial equation [A] implicitly:

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

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

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x}{2 y} = \frac{x}{y}$ ..... [B]

We now differentiate this second equation [B] implicitly and apply the quotient rule, giving:

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

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

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{y - x \frac{\mathrm{dy}}{\mathrm{dx}}}{y} ^ 2$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{y - x \left(\frac{x}{y}\right)}{y} ^ 2 \setminus \setminus \setminus$ using [B]

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{\left({y}^{2} - {x}^{2}\right) \frac{1}{y}}{y} ^ 2$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{{y}^{2} - {x}^{2}}{y} ^ 3$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = - \frac{25}{y} ^ 3 \setminus \setminus \setminus$ using [A] QED