Question

If $y^2=P\left(x\right)$ is a polynomial of degree 3, then $2\left(\frac{d}{dx}\right)(y^3.\frac{d^{2}y}{d{x}^2})$ 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. A constant

Answer 1

Answer: (C)

$P\left(x\right).P^{\prime \prime \prime }\left(x\right)$

$y^2=P\left(x\right)$
$P\left(x\right)$ is a polynomial of degree 3
$y^2=P\left(x\right)$
$2y\frac{dy}{dx}=P^{{}^{\prime }}\left(x\right)$
$2y\frac{d^{2}y}{d{x}^2}+2{\left(\frac{dy}{dx}\right)}^2=P^{{}^{\prime \prime }}\left(x\right)$
$\frac{d^{2}y}{d{x}^2}=\frac{P^{{}^{\prime \prime }}\left(x\right)-2{\left[\frac{P^{{}^{\prime }}\left(x\right)}{2y}\right]}^2}{2y}$
$y^3\frac{d^{2}y}{d{x}^2}=\frac{y^3{P}^{{}^{\prime \prime }}\left(x\right)}{2y}-\frac{y^3{\left[P^{{}^{\prime }}\left(x\right)\right]}^2}{4y^3}$
$=\frac{y^2{P}^{{}^{\prime \prime }}\left(x\right)}{2}-\frac{{\left[P^{{}^{\prime }}\left(x\right)\right]}^2}{4}$
$\left[y^3\frac{d^{2}y}{d{x}^2}\right]=\frac{P\left(x\right)P^{{}^{\prime \prime }}\left(x\right)}{2}-\frac{{\left[P^{{}^{\prime }}\left(x\right)\right]}^2}{4}$
$2\frac{d}{dx}\left[y^3\frac{d^{2}y}{d{x}^2}\right]=P^{{}^{\prime }}\left(x\right)P^{{}^{\prime \prime }}\left(x\right)+P\left(x\right)P^{{}^{\prime \prime \prime }}\left(x\right)-P^{{}^{\prime }}\left(x\right)P^{{}^{\prime \prime }}\left(x\right)$
$=P\left(x\right)P^{{}^{\prime \prime \prime }}\left(x\right)$