If frac{d^2x}{dy^2} left( frac{dy}{dx} right) | vedclass

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(D) We know that $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$.By differentiating both sides with respect to $x$,we get:$\frac{d^2y}{dx^2} = \frac{d}{dx}

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(D) We know that $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$.By differentiating both sides with respect to $x$,we get:$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \left( \frac{dx}{dy} \right)^{-1} \right)$.Using the chain rule,$\frac{d^2y}{dx^2} = -\left( \frac{dx}{dy} \right)^{-2} \cdot \frac{d}{dx} \left( \frac{dx}{dy} \right)$.Since $\frac{d}{dx} = \frac{dy}{dx} \cdot \frac{d}{dy}$,we have:$\frac{d^2y}{dx^2} = -\left( \frac{dx}{dy} \right)^{-2} \cdot \frac{dy}{dx} \cdot \frac{d}{dy} \left( \frac{dx}{dy} \right)$.Substituting $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$,we get:$\frac{d^2y}{dx^2} = -\left( \frac{dx}{dy} \right)^{-2} \cdot \left( \frac{dx}{dy} \right)^{-1} \cdot \frac{d^2x}{dy^2} = -\left( \frac{dx}{dy} \right)^{-3} \cdot \frac{d^2x}{dy^2}$.Since $\frac{dx}{dy} = \left( \frac{dy}{dx} \right)^{-1}$,then $\left( \frac{dx}{dy} \right)^{-3} = \left( \frac{dy}{dx} \right)^3$.Thus,$\frac{d^2y}{dx^2} = -\left( \frac{dy}{dx} \right)^3 \cdot \frac{d^2x}{dy^2}$.Rearranging the terms,we get $\frac{d^2x}{dy^2} \left( \frac{dy}{dx} \right)^3 + \frac{d^2y}{dx^2} = 0$.Comparing this with the given equation,we find $K = 0$.

If $\frac{d^2x}{dy^2} \left( \frac{dy}{dx} \right)^3 + \frac{d^2y}{dx^2} = K$,then the value of $K$ is equal to

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