If x^{2}+y^{2}=1,then frac{d^{2} x}{d y^{2}}= | vedclass

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(C) Given the equation $x^{2}+y^{2}=1$.Differentiating both sides with respect to $y$,we get:$2x \frac{dx}{dy} + 2y = 0$$2x \frac{dx}{dy} = -2y$$\frac

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$-\frac{1}{x^{3}}$

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(C) Given the equation $x^{2}+y^{2}=1$.Differentiating both sides with respect to $y$,we get:$2x \frac{dx}{dy} + 2y = 0$$2x \frac{dx}{dy} = -2y$$\frac{dx}{dy} = -\frac{y}{x}$Now,differentiating $\frac{dx}{dy}$ with respect to $y$ using the quotient rule:$\frac{d^{2}x}{dy^{2}} = \frac{d}{dy} \left( -\frac{y}{x} \right) = -\left[ \frac{x(1) - y(\frac{dx}{dy})}{x^{2}} \right]$Substitute $\frac{dx}{dy} = -\frac{y}{x}$ into the expression:$\frac{d^{2}x}{dy^{2}} = -\left[ \frac{x - y(-\frac{y}{x})}{x^{2}} \right]$$\frac{d^{2}x}{dy^{2}} = -\left[ \frac{x + \frac{y^{2}}{x}}{x^{2}} \right] = -\left[ \frac{x^{2} + y^{2}}{x^{3}} \right]$Since $x^{2}+y^{2}=1$,we have:$\frac{d^{2}x}{dy^{2}} = -\frac{1}{x^{3}}$

If $x^{2}+y^{2}=1$,then $\frac{d^{2} x}{d y^{2}}=$

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