Question

If $y^2=p(x),$ 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"'(x) + p' (x)
• B. p"(x) p'" (x)
• C. p(x) p"' (x)
• D. a constant

Answer 1

Answer: (C)

$y^2=p\left(x\right)\Rightarrow 2y\frac{dy}{dx}=p^{\prime }\left(x\right)$
$\Rightarrow 2y\frac{d^{2}y}{d{x}^2}+2{\left(\frac{dy}{dx}\right)}^2=p"\left(x\right)$
$\therefore 2y\frac{d^{3}y}{d{x}^3}+2\frac{dy}{dx}.\frac{d^{2}y}{d{x}^2}+4\frac{dy}{dx}.\frac{d^{2}y}{d{x}^2}=p^{\prime \prime \prime }\left(x\right)$
$\Rightarrow 2y{y}_3+6y_1{y}_2=p^{\prime \prime \prime }\left(x\right)$
Now $2\frac{d}{dx}\left(y^3{y}_2\right)=2\left[y^3{y}_3+3y^2{y}_1{y}_2\right]$
$=y^2\left[2y{y}_3+6y_1{y}_2\right]=y^2{p}^{\prime \prime \prime }\left(x\right)=p\left(x\right)p^{\prime \prime \prime }\left(x\right)$