Question

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

• A. $p^{\prime \prime \prime }\left(x\right)+p^{\prime }\left(x\right)$
• B. $p^{\prime \prime }\left(x\right)+p^{\prime \prime \prime }\left(x\right)$
• C. $p\left(x\right)p^{\prime \prime \prime }\left(x\right)$
• D. constant

Answer 1

Answer: (C)

$p\left(x\right)p^{\prime \prime \prime }\left(x\right)$

$y=\sqrt{p\left(x\right)}$
$\frac{dy}{dx}=\frac{1}{2}\frac{p^{\prime }\left(x\right)}{\sqrt{p\left(x\right)}}$
$\frac{d^{2}y}{d{x}^2}=\frac{1}{2}\left(\frac{p^{\prime \prime }\left(x\right)\sqrt{p\left(x\right)}-\frac{\left(p^{\prime }\right(x){)}^2}{2\sqrt{p\left(x\right)}}}{\left(p\right(x\left)\right)}\right)$
$y^3\frac{d^{2}y}{d{x}^2}=\frac{2p^{{}^{\prime \prime }}\left(x\right)p\left(x\right)-\left(p^{{}^{\prime }}\right(x){)}^2}{4}$
$2\frac{d}{dx}\left(y^3\frac{d^{2}y}{d{x}^2}\right)=\frac{1}{2}\left(2p^{\prime \prime \prime }\right(x\left)p\right(x)+2p^{\prime \prime }(x\left)p^{\prime }\right(x)-2p^{\prime }(x\left)p^{\prime \prime }\right(x\left)\right)$
$2\frac{d}{dx}\left(y^3\frac{d^{2}y}{d{x}^2}\right)=p^{{}^{\prime \prime \prime }}\left(x\right)p\left(x\right)$