If {y^2} = p(x) is a polynomial of degree thr | vedclass

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(C) Given ${y^2} = p(x)$.Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = p'(x)$.Thus,$\frac{dy}{dx} = \frac{p'(x)}{2y}$.Differentiating

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$p(x).p'''(x)$

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(C) Given ${y^2} = p(x)$.Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = p'(x)$.Thus,$\frac{dy}{dx} = \frac{p'(x)}{2y}$.Differentiating again with respect to $x$,we get $2 \left( \frac{dy}{dx} \right)^2 + 2y \frac{d^2y}{dx^2} = p''(x)$.Substituting $\frac{dy}{dx} = \frac{p'(x)}{2y}$,we have $2 \left( \frac{p'(x)}{2y} \right)^2 + 2y \frac{d^2y}{dx^2} = p''(x)$.$2y \frac{d^2y}{dx^2} = p''(x) - \frac{(p'(x))^2}{2y^2} = p''(x) - \frac{(p'(x))^2}{2p(x)}$.Multiplying by $y^2$,we get $2y^3 \frac{d^2y}{dx^2} = p(x) p''(x) - \frac{1}{2} (p'(x))^2$.Now,differentiating $2y^3 \frac{d^2y}{dx^2}$ with respect to $x$:$\frac{d}{dx} \left( 2y^3 \frac{d^2y}{dx^2} \right) = \frac{d}{dx} \left( p(x) p''(x) - \frac{1}{2} (p'(x))^2 \right)$.$= p'(x) p''(x) + p(x) p'''(x) - \frac{1}{2} \cdot 2 p'(x) p''(x)$.$= p'(x) p''(x) + p(x) p'''(x) - p'(x) p''(x) = p(x) p'''(x)$.

If ${y^2} = p(x)$ is a polynomial of degree three,then $2{d \over {dx}}\left\{ {{y^3}.{{{d^2}y} \over {d{x^2}}}} \right\} =$

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