If y^2 = P(x) is a polynomial of degree 3,the | vedclass

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(C) Given $y^2 = P(x)$. Differentiating with respect to $x$:$2y y_1 = P'(x) \implies y y_1 = \frac{1}{2} P'(x)$.Differentiating again:$y y_2 + y_1^2 =

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$P(x) \cdot P'''(x)$

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(C) Given $y^2 = P(x)$. Differentiating with respect to $x$:$2y y_1 = P'(x) \implies y y_1 = \frac{1}{2} P'(x)$.Differentiating again:$y y_2 + y_1^2 = \frac{1}{2} P''(x)$.Multiply by $2y^2$:$2y^3 y_2 + 2y^2 y_1^2 = y^2 P''(x) = P(x) P''(x)$.Since $y_1 = \frac{P'(x)}{2y}$,then $y_1^2 = \frac{(P'(x))^2}{4y^2} = \frac{(P'(x))^2}{4P(x)}$.Substituting $y_1^2$:$2y^3 y_2 + 2P(x) \cdot \frac{(P'(x))^2}{4P(x)} = P(x) P''(x) \implies 2y^3 y_2 + \frac{1}{2} (P'(x))^2 = P(x) P''(x)$.Thus,$2y^3 y_2 = P(x) P''(x) - \frac{1}{2} (P'(x))^2$.Differentiating both sides with respect to $x$:$\frac{d}{dx} (2y^3 y_2) = P'(x) P''(x) + P(x) P'''(x) - P'(x) P''(x) = P(x) P'''(x)$.Therefore,the expression equals $P(x) P'''(x)$.

If $y^2 = P(x)$ is a polynomial of degree $3$,then $2 \left( \frac{d}{dx} \right) \left( y^3 \frac{d^2y}{dx^2} \right)$ equals :

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