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"' (x) + p' x
• B. p''(x). p"'(x)
• C. p (x). p"' (x)
• D. None of these

Answer 1

The correct option is C p (x). p"' (x)

$y^2=P\left(x\right)\Rightarrow 2y{y}^{\prime }=p^{\prime }\left(x\right)$-----(i)
$\Rightarrow \left(2y\right)y^{\prime \prime }+y^{\prime }\left(2y^{\prime }\right)=p^{\prime \prime }\left(x\right)$
$\Rightarrow 2y{y}^{\prime \prime }=p^{\prime \prime }\left(x\right)-2(y^{\prime }{)}^2$
$\Rightarrow 2y^3{y}^{\prime \prime }=y^2{p}^{\prime \prime }\left(x\right)-2(y{y}^{\prime }{)}^2$
$\Rightarrow 2y^3{y}^{\prime \prime }=y^2{p}^{\prime \prime }\left(x\right)-2\frac{\left(p^{\prime }\right(x{)}^2)}{4}[fromEq.(i\left)\right]$
$\Rightarrow 2y^3{y}^{\prime \prime }=p\left(x\right)p^{\prime \prime }\left(x\right)-\frac{1}{2}{p^{\prime }\left(x\right)}^2$
$\therefore \frac{d}{dx}\left(2y^3{y}^{\prime \prime }\right)=p\left(x\right)p^{\prime \prime \prime }\left(x\right)+p^{\prime }\left(x\right)p^{\prime \prime }\left(x\right)-p^{\prime }\left(x\right)p^{\prime \prime }\left(x\right)=p\left(x\right).p^{\prime \prime \prime }\left(x\right)$
$\Rightarrow 2\frac{d}{dx}\left(y^3\frac{d^{2}y}{d{x}^2}\right)=p\left(x\right)p^{\prime \prime \prime }\left(x\right)$