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

From $y^2=P\left(x\right)$, we have $2y{y}_1=P^{\prime }\left(x\right)$, i.e.,
$2y_1=P^{\prime }\left(x\right)/y$
$\Rightarrow 2y_2=\frac{y{P}^{\prime \prime }\left(x\right)-P^{\prime }\left(x\right)y_1}{y^2}$
$=\frac{y{P}^{\prime \prime }\left(x\right)-P^{\prime }\left(x\right).P^{\prime }\left(x\right)/2y}{y^2}$
$=\frac{2y^2{P}^{\prime \prime }\left(x\right)-\left(P^{\prime }\right(x){)}^2}{2y^3}$
$\Rightarrow 2y_2{y}^3=\frac{1}{2}\left[2P\right(x\left)P^{\prime \prime }\right(x)-(P^{\prime }\left(x\right){)}^2]$
Now, differentiating w.r.t. x on both sides,
$\Rightarrow 2\frac{d}{dx}\left(y^3\frac{d^{2}y}{d{x}^2}\right)=\frac{1}{2}[2\left\{P^{\prime }\right(x\left)P^{\prime \prime }\right(x)+P(x\left)P^{\prime \prime \prime }\right(x\left)\right\}-2P^{\prime }(x\left)P^{\prime \prime }\right(x\left)\right]$
$=\frac{1}{2}.2P\left(x\right)P^{\prime \prime \prime }\left(x\right)=P\left(x\right)P^{\prime \prime \prime }\left(x\right)$