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

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

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(c) : Given $y^{2}=P$ Differentiating $2 y y^{\prime}=P^{\prime}$\ $	herefore \quad\left(y^{\prime}ight)^{2}=\frac{\left(P^{\prime}ight)^{2}}{4 y^{2}}=\frac{\left(P^{\prime}ight)^{2}}{4 P}$\ Differentiating again\ $2 y y^{\prime \prime}+2\left(y^{\prime}ight)^{2}=P^{\prime \prime}$\ $	herefore \quad 2 y y^{\prime \prime}=P^{\prime \prime}-2\left(y^{\prime}ight)^{2}$\ Using (1),\ $2 y y^{\prime \prime}=P^{\prime \prime}-2\left[\frac{\left(P^{\prime}ight)^{2}}{4 P}ight]=P^{\prime \prime}-\frac{\left(P^{\prime}ight)^{2}}{2 P}$\ $	herefore \quad 2 P y y^{\prime \prime}=P P^{\prime \prime}-\frac{\left(P^{\prime}ight)^{2}}{2}$\ Now $P=y^{2}$\ $	herefore \quad 2 y^{3} y^{\prime \prime}=P P^{\prime \prime}-\frac{\left(P^{\prime}ight)^{2}}{2}$\ Differentiating again\ $2 \frac{d}{d x}\left(y^{3} y^{\prime \prime}ight)=P P^{\prime \prime \prime}+P^{\prime} P^{\prime \prime}-P^{\prime} P^{\prime \prime}$\ or $2 \frac{d}{d x}\left(y^{3} y^{\prime \prime}ight)=P P^{\prime \prime \prime}$.\