If y=f(x) is a thrice differentiable function | vedclass

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(D) Given,$y=f(x)$ is a differentiable and bijective function. We know that $\frac{d x}{d y} = \frac{1}{\frac{d y}{d x}} = \left(\frac{d y}{d x}\right

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(D) Given,$y=f(x)$ is a differentiable and bijective function. We know that $\frac{d x}{d y} = \frac{1}{\frac{d y}{d x}} = \left(\frac{d y}{d x}\right)^{-1}$. Differentiating both sides with respect to $y$: $\frac{d^2 x}{d y^2} = \frac{d}{d y} \left[ \left(\frac{d y}{d x}\right)^{-1} \right]$. Using the chain rule: $\frac{d^2 x}{d y^2} = \frac{d}{d x} \left[ \left(\frac{d y}{d x}\right)^{-1} \right] \cdot \frac{d x}{d y}$. $\frac{d^2 x}{d y^2} = -\left(\frac{d y}{d x}\right)^{-2} \cdot \frac{d^2 y}{d x^2} \cdot \frac{1}{\frac{d y}{d x}}$. $\frac{d^2 x}{d y^2} = -\frac{\frac{d^2 y}{d x^2}}{\left(\frac{d y}{d x}\right)^3}$. Multiplying both sides by $\left(\frac{d y}{d x}\right)^3$: $\frac{d^2 x}{d y^2} \cdot \left(\frac{d y}{d x}\right)^3 = -\frac{d^2 y}{d x^2}$. Therefore,$\frac{d^2 x}{d y^2} \left(\frac{d y}{d x}\right)^3 + \frac{d^2 y}{d x^2} = 0$.

If $y=f(x)$ is a thrice differentiable function and a bijection,then $\frac{d^2 x}{d y^2}\left(\frac{d y}{d x}\right)^3+\frac{d^2 y}{d x^2}=$

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