Question

Hello
plz help me in this.​

Answer 1

Answer:

$y = \sqrt{ \frac{1 - x}{1 + x} } \\ \\ {y}^{2} = \frac{1 - x}{1 + x} \\ \\ differnting \: \: on \: \: bothsides \\ \\ 2y \frac{dy}{dx} = \frac{(1 + x)( - 1) \: - (1 - x)(1) }{ {(1 + x)}^{2} } \\ \\ \frac{dy}{dx} = \frac{ - 2 }{ {2y(1 + x)}^{2} } \\ \\ {(1 - x ) }^{2} \frac{dy}{dx} + y \: = 0$

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