Question

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

• 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(X)

$\Rightarrow$ y= $\surd \left(P\right(x\left)\right)$

$\Rightarrow$ $y^3$ = ${P\left(x\right)}^{3/2}$

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

$\frac{d^{2}y}{d{x}^2}$ = $\frac{1}{2}$ $\frac{\sqrt{\left(P\right(x\left)\right).P^{\prime \prime }\left(x\right)-({P^{\prime }\left(x\right))}^2}/2\sqrt{\left(P\right(x\left)\right)}}{P\left(x\right)}$ -------(2)

$y^3$.$\frac{d^{2}y}{d{x}^2}$ = ${P\left(x\right)}^{3/2}$. $\frac{1}{2}$ $\frac{\sqrt{\left(P\right(x\left)\right).P^{\prime \prime }\left(x\right)}-({P^{\prime }\left(x\right))}^2/2\sqrt{\left(P\right(x\left)\right)}}{P\left(x\right)}$ (From (2))

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

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

)

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

So, option c is correct.