Question

If $y^2=P\left(x\right)$ be a cubic polynomial, 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

The correct option is C $P\left(x\right)P^{\prime \prime \prime }\left(x\right)$

$y^2=P\left(x\right)$

$P^{\prime }\left(x\right)=2y{y}^{\prime }$

$P^{\prime \prime }\left(x\right)=2y{y}^{\prime \prime }+2y^{\prime 2}$
$\Rightarrow P^{\prime \prime \prime }\left(x\right)=2y{y}^{\prime \prime \prime }+6y^{\prime }{y}^{\prime \prime }$
$\therefore 2\frac{d}{dx}\left(y^3{y}^{\prime \prime }\right)=2[y^3{y}^{\prime \prime \prime }+3y^2{y}^{\prime }{y}^{\prime \prime }]$
$=y^2[2y{y}^{\prime \prime \prime }+6y^{\prime }{y}^{\prime \prime }]$
$=P\left(x\right)P^{\prime \prime \prime }\left(x\right)$.