If x=y1-y2 then dydx=

x=y 1-y2 dxdy=y (-2y)21-y2+1-y2 =-y2+(1-y2)1-y2=1-2y21-y2 dydx=1-y21-2y2

If x=y1-y2 then dydx=

x=y 1-y2 dxdy=y (-2y)21-y2+1-y2 =-y2+(1-y2)1-y2=1-2y21-y2 dydx=1-y21-2y2

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$If x = y \sqrt{1 - y^{2}} then \frac{dy}{dx} =$

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$\frac{\sqrt{1 + y^{2}}}{1 - 2 y^{2}}$

$\frac{\sqrt{1 - y^{2}}}{1 - 2 y^{2}}$

$\frac{\sqrt{1 - y^{2}}}{1 + 2 y^{2}}$

$\frac{\sqrt{1 + y^{2}}}{1 + 2 y^{2}}$

answer is B.

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Detailed Solution

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

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$If x = y \sqrt{1 - y^{2}} then \frac{dy}{dx} =$