frac{d^2 x}{d y^2} equals | vedclass

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(D) We know that $\frac{d x}{d y} = \frac{1}{d y / d x} = \left(\frac{d y}{d x}\right)^{-1}$.Now,differentiate both sides with respect to $y$:$\frac{d

Vedclass · Accepted Answer

$-\left(\frac{d^2 y}{d x^2}\right)\left(\frac{d y}{d x}\right)^{-3}$

Vedclass · Answer

(D) We know that $\frac{d x}{d y} = \frac{1}{d y / d x} = \left(\frac{d y}{d x}\right)^{-1}$.Now,differentiate 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,we have:$\frac{d^2 x}{d y^2} = \frac{d}{d x} \left[ \left(\frac{d y}{d x}\right)^{-1} \right] \cdot \frac{d x}{d y}$.Applying the power rule:$\frac{d^2 x}{d y^2} = -\left(\frac{d y}{d x}\right)^{-2} \cdot \frac{d^2 y}{d x^2} \cdot \frac{d x}{d y}$.Since $\frac{d x}{d y} = \left(\frac{d y}{d x}\right)^{-1}$,we substitute this into the expression:$\frac{d^2 x}{d y^2} = -\left(\frac{d y}{d x}\right)^{-2} \cdot \frac{d^2 y}{d x^2} \cdot \left(\frac{d y}{d x}\right)^{-1}$.Combining the powers of $\frac{d y}{d x}$:$\frac{d^2 x}{d y^2} = -\left(\frac{d^2 y}{d x^2}\right) \left(\frac{d y}{d x}\right)^{-3}$.

$\frac{d^2 x}{d y^2}$ equals

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