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 $x$:$2y y' = p'(x)$Differentiating again:$2y y'' + 2(y')^2 = p''(x)$Differentiating a third tim

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

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(C) Given $y^2 = p(x)$.Differentiating with respect to $x$:$2y y' = p'(x)$Differentiating again:$2y y'' + 2(y')^2 = p''(x)$Differentiating a third time:$2y y''' + 2y' y'' + 4y' y'' = p'''(x)$$2y y''' + 6y' y'' = p'''(x)$Now,consider the expression $E = 2\frac{d}{dx}\left( y^3 y'' \right)$:$E = 2(3y^2 y' y'' + y^3 y''')$$E = 2y^2(3y' y'' + y y''')$From the third derivative equation,we have $6y' y'' + 2y y''' = p'''(x)$,which implies $3y' y'' + y y''' = \frac{1}{2} p'''(x)$.Substituting this into the expression for $E$:$E = 2y^2 \left( \frac{1}{2} p'''(x) \right)$$E = y^2 p'''(x)$Since $y^2 = p(x)$,we get:$E = p(x) p'''(x)$.

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

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