Question

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

• A. P'''(x)+P'(x)
• B. P''(x).P'''(x)
• C. P(x).P'''(x)
• D. a constant

Answer 1

The correct option is C P(x).P'''(x)

$y^2=P\left(x\right)\Rightarrow 2y\frac{dy}{dx}=p^{\prime }\left(x\right)Or2{\left(\frac{dy}{dx}\right)}^2+2y\frac{d^{2}y}{d{x}^2}=P^{\prime \prime }\left(x\right)Or2y\frac{d^{2}y}{d{x}^2}=P^{\prime \prime }-2{\left(\frac{dy}{dx}\right)}^2=P^{\prime \prime }-\frac{P^{{}^{\prime }2}}{2y^2}\because 2y^3\frac{d^{2}y}{d{x}^2}=y^2{p}^{\prime \prime }-\frac{1}{2}{P}^{{}^{\prime }2}=P{P}^{\prime \prime }=\frac{1}{2}{P}^{{}^{\prime }2}\because 2\frac{d}{dx}\left(y^3\frac{d^{2}y}{d{x}^2}\right)=P^{\prime }{P}^{\prime \prime }+P{P}^{\prime \prime \prime }-P^{\prime }{P}^{\prime \prime }=P{P}^{\prime \prime \prime }$