Question

X is equals to Y of root 1 — y square then DY /DX through derivations






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Answer 1

Answer:

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Answer 2

$x = y \sqrt{1 - {y}^{2} } \\diff \: w.r.t.x \\ 1 = y \times \frac{1}{2 \sqrt{1 - {y}^{2} } } \times - 2y \times \frac{dy}{dx} \\ \: \: \: \: \: \: + \sqrt{1 - {y}^{2} } \frac{dy}{dx} \\ 1 = \frac{dy}{dx}( \frac{ { - y}^{2} + 1 - {y}^{2} }{ \sqrt{1 - {y}^{2} } } ) \\ \ \frac{ \sqrt{1 - {y}^{2} } }{1 - 2 {y}^{2} } = \frac{dy}{dx}$