If y2 = p(x), a polynomial of degree 3 then 2 d/(d x)≤ft(y^3 (d^2 y)/…

$y^2=p(x) \Rightarrow 2 y y_1=p'(x)$

$\Rightarrow 2\left(y y_2+y_1 y_1\right)=p''(x)$

$\Rightarrow y y_2=\frac{1}{2}\left(p''(x)-2 y_1{ }^2\right)$

$\Rightarrow y^3 y_2=\frac{1}{2}\left(p''(x) y^2-2\left(y y_1\right)^2\right)$

$=\frac{1}{2} p''(x) . p(x)-\left(\frac{1}{2} p'(x)\right)^2$

$\Rightarrow \frac{d}{d x}\left(y^3 y_2\right)=\frac{1}{2}\left[p'''(x) p(x)+p'(x) p''(x)\right]-2 \times\left(\frac{1}{2} p'(x)\right) \times\left(\frac{1}{2} p''(x)\right)$

$\Rightarrow 2 \frac{d}{d x} \left(y^3 y_2\right)=p(x) p'''(x)$

Hence (3) is correct answer.