# If y=root1-x/1+x.then prove that (1-X)^2dy/dx+y=0?

## If $y = \sqrt{\frac{1 - x}{1 + x}}$, then prove that $\left(1 - {x}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}} + y = 0$

Apr 13, 2018

#### Explanation:

$y = \sqrt{\frac{1 - x}{1 + x}}$

$y = \sqrt{\frac{1 - x}{1 + x} \times \frac{1 - x}{1 - x}}$

y=sqrt((1-x)^2/(1-x^2)

$y = \frac{1 - x}{\sqrt{1 - {x}^{2}}} \ldots \to \left(A\right)$

$\text{Using "color(blue)"Quotient Rule}$

(dy)/(dx)=((sqrt(1-x^2))(0-1)-(1-x)(1/(2sqrt(1-x^2)) (-2x)))/((sqrt(1-x^2))^2)

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

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

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

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

$\left(1 - {x}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1 - x}{\sqrt{1 - {x}^{2}}} \ldots \to U s e \left(A\right)$

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

$\left(1 - {x}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}} + y = 0$