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

Limits and Derivatives

Solution:

$y^{2} =p\left(x\right) \Rightarrow 2y \frac{dy}{dx} =p'\left(x\right)$
$\Rightarrow 2y \frac{d^{2}y}{dx^{2}} +2 \left(\frac{dy}{dx}\right)^{2} =p"\left(x\right)$
$ \therefore 2y \frac{d^{3}y}{dx^{3}} + 2 \frac{dy}{dx}. \frac{d^{2}y}{dx^{2}}+ 4 \frac{dy}{dx}. \frac{d^{2}y}{dx^{2}} = p''' \left(x\right)$
$ \Rightarrow 2yy_{3} + 6y_{1}y_{2} = p'''\left(x\right) $
Now $2 \frac{d}{dx} \left(y^{3}y_{2}\right) = 2 \left[y^{3}y_{3} + 3y^{2}y_{1}y_{2}\right] $
$=y^{2}\left[2yy_{3} + 6y_{1}y_{2}\right] = y^{2} p'''\left(x\right) = p\left(x\right)p'''\left(x\right) $