# Find d²y/dx² in terms of t if x= (1-t²)/(1+t²) and y=t/(1+t²)?

Aug 25, 2015

$\frac{- 4 {y}^{2} - {x}^{2}}{16 {y}^{3}}$

#### Explanation:

First find $\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{\left(1 + {t}^{2}\right) \left(- 2 t\right) - \left(1 - {t}^{2}\right) \left(2 t\right)}{1 + {t}^{2}} ^ 2$=$\frac{- 4 t}{1 + {t}^{2}} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{1 \left(1 + {t}^{2}\right) - t \left(2 t\right)}{1 + {t}^{2}} ^ 2$=$\frac{1 - {t}^{2}}{1 + {t}^{2}} ^ 2$

This would give $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - {t}^{2}}{- 4 t}$

Since $y \left(1 + {t}^{2}\right) = t$ and $x \left(1 + {t}^{2}\right) = 1 - {t}^{2}$, you get:

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

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{- 4 y - x \left(- 4\right) \frac{\mathrm{dy}}{\mathrm{dx}}}{16 {y}^{2}}$=$\frac{- 4 y + 4 x \frac{\mathrm{dy}}{\mathrm{dx}}}{16 {y}^{2}}$

=$\frac{- 4 {y}^{2} - {x}^{2}}{16 {y}^{3}}$

Aug 28, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{- 4 y}$

#### Explanation:

It is like this:

It was derived that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - {t}^{2}}{- 4 t}$.
Now, from the given equations, it is evident that $1 - {t}^{2} = x \left(1 + {t}^{2}\right)$ and
$t = y \left(1 + {t}^{2}\right)$. Hence,

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