Question

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

• A. p"'(x) + p' (x)
• B. p"(x) p'" (x)
• C. p(x) p"' (x)
• D. a constant

Answer 1

The Correct Option is C

$y^{2} =p\left(x\right) \Rightarrow 2y \frac{dy}{dx} =p'\left(x\right)$ $\Rightarrow 2y \frac{d^{2}y}{dx^{2}} +2 \left(\frac{dy}{dx}\right)^{2} =p"\left(x\right)$ $ \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) $