If x = ysqrt{1 - y^2},then frac{dy}{dx} = | vedclass

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(C) Given equation: $x = y\sqrt{1 - y^2}$Differentiating both sides with respect to $x$ using the product rule and chain rule:$1 = \frac{dy}{dx} \cdot

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

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(C) Given equation: $x = y\sqrt{1 - y^2}$Differentiating both sides with respect to $x$ using the product rule and chain rule:$1 = \frac{dy}{dx} \cdot \sqrt{1 - y^2} + y \cdot \frac{1}{2\sqrt{1 - y^2}} \cdot (-2y) \cdot \frac{dy}{dx}$Simplify the expression:$1 = \frac{dy}{dx} \left( \sqrt{1 - y^2} - \frac{y^2}{\sqrt{1 - y^2}} \right)$Find a common denominator inside the bracket:$1 = \frac{dy}{dx} \left( \frac{1 - y^2 - y^2}{\sqrt{1 - y^2}} \right)$$1 = \frac{dy}{dx} \left( \frac{1 - 2y^2}{\sqrt{1 - y^2}} \right)$Solving for $\frac{dy}{dx}$:$\frac{dy}{dx} = \frac{\sqrt{1 - y^2}}{1 - 2y^2}$

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

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