Question

If $y^2=P\left(x\right)$ is 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. $P\left(x\right)+P^{\prime }\left(x\right)$
• B. $P\left(x\right)P^{\prime }\left(x\right)$
• C. $P\left(x\right)P^{\prime \prime \prime }\left(x\right)$
• D. a constant

Answer 1

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

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

Differentiating both sides w.r.t $x$ we get

$2y{y}_1=P^{\prime }\left(x\right)\Rightarrow 2y_1=\frac{P^{\prime }\left(x\right)}{y}$

Again differentiating w.r.t $x$, we get

$2y_2=\frac{y{P}^{\prime \prime }\left(x\right)-P^{\prime }\left(x\right)y_1}{y^2}=\frac{y{P}^{\prime \prime }\left(x\right)-\frac{P^{\prime }\left(x\right).P^{\prime }\left(x\right)}{2y}}{y^2}$

$=\frac{2y^2{P}^{\prime \prime }\left(x\right)-{\left(P^{\prime }\left(x\right)\right)}^2}{2y^3}=\frac{2P\left(x\right)P^{\prime \prime }\left(x\right)-{\left(P^{\prime }\left(x\right)\right)}^2}{2y^3}$

$\Rightarrow 2y_2{y}^3=\frac{1}{2}[2P\left(x\right)P^{\prime \prime }\left(x\right)-{\left(P^{\prime }\left(x\right)\right)}^2]$

$\Rightarrow 2\frac{d}{dx}\left(y^3\frac{d^{2}y}{d{x}^2}\right)=\frac{1}{2}[2(P^{\prime }\left(x\right)P^{\prime \prime }\left(x\right)+P\left(x\right)P^{\prime \prime \prime }\left(x\right))-2P^{\prime }\left(x\right)P^{\prime \prime }\left(x\right)]$

$=\frac{1}{2}\left(2P\left(x\right)P^{\prime \prime \prime }\left(x\right)\right)=P\left(x\right)P^{\prime \prime \prime }\left(x\right)$