Question

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

• A. $P\text{'}\text{'}\text{'}\left(x\right)+P\text{'}\left(x\right)$
• B. $P\text{'}\text{'}\left(x\right)P\text{'}\text{'}\text{'}\left(x\right)$
• C. $P\left(x\right)P\text{'}\text{'}\text{'}\left(x\right)$
• D. a constant

Answer 1

The correct option is C

$P\left(x\right)P\text{'}\text{'}\text{'}\left(x\right)$

Explanation for the correct option.

Step 1. Differentiate the given equation twice with respect to $x$.

Differentiate the equation $y^2=P\left(x\right)$ with respect to $x$.

$$\begin{gathered} 2yy\text{'}=P\text{'}\left(x\right) \\ \Rightarrow y\text{'}=\frac{P\text{'}\left(x\right)}{2y}...\left(1\right) \end{gathered}$$

Now again differentiate $2yy\text{'}=P\text{'}\left(x\right)$ with respect to $x$.

$$\begin{gathered} 2(y\text{'})(y\text{'})+2yy\text{'}\text{'}=P\text{'}\text{'}\left(x\right) \\ \Rightarrow 2yy\text{'}\text{'}=P\text{'}\text{'}\left(x\right)-2{(y\text{'})}^2 \\ \Rightarrow 2yy\text{'}\text{'}=P\text{'}\text{'}\left(x\right)-2{\left(\frac{P\text{'}\left(x\right)}{2y}\right)}^2\left[\mathbf{Equation}\mathbf{}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{}\mathit{y}\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{P}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{2}\mathbf{y}}\right] \\ \Rightarrow 2yy\text{'}\text{'}=P\text{'}\text{'}\left(x\right)-\frac{{\left\{P\text{'}\left(x\right)\right\}}^2}{2y^2} \end{gathered}$$

Now multiply both sides of the equation by $y^2$.

$$\begin{gathered} 2yy\text{'}\text{'}\times y^2=\left[P\text{'}\text{'}\left(x\right)-\frac{{\left\{P\text{'}\left(x\right)\right\}}^2}{2y^2}\right]\times y^2 \\ \Rightarrow 2y^{3}y\text{'}\text{'}=y^{2}P\text{'}\text{'}\left(x\right)-\frac{{\left\{P\text{'}\left(x\right)\right\}}^2}{2} \\ \Rightarrow 2y^{3}y\text{'}\text{'}=P\left(x\right)P\text{'}\text{'}\left(x\right)-\frac{{\left\{P\text{'}\left(x\right)\right\}}^2}{2}\left[{\mathit{y}}^{\mathbf{2}}\mathbf{=}\mathit{P}\mathbf{\left(}\mathit{x}\mathbf{\right)}\right] \end{gathered}$$

So, $2y^3\frac{d^{2}y}{d{x}^2}=P\left(x\right)P\text{'}\text{'}\left(x\right)-\frac{{\left\{P\text{'}\left(x\right)\right\}}^2}{2}$.

Step 2. Differentiate the equation with respect to $x$ to find the value of $2\frac{d}{dx}\left(y^3\frac{d^{2}y}{d{x}^2}\right)$.

Now differentiate the equation $2y^3\frac{d^{2}y}{d{x}^2}=P\left(x\right)P\text{'}\text{'}\left(x\right)-\frac{{\left\{P\text{'}\left(x\right)\right\}}^2}{2}$ with respect to $x$.

$\begin{array}{rcl}2\frac{d}{dx}\left(y^3\frac{d^{2}y}{d{x}^2}\right)& =& \frac{d}{dx}\left(P\left(x\right)P\text{'}\text{'}\left(x\right)-\frac{{\left\{P\text{'}\left(x\right)\right\}}^2}{2}\right)\\ & =& P\text{'}\left(x\right)P\text{'}\text{'}\left(x\right)+P\left(x\right)P\text{'}\text{'}\text{'}\left(x\right)-\frac{2P\text{'}\left(x\right)P\text{'}\text{'}\left(x\right)}{2}\\ & =& P\text{'}\left(x\right)P\text{'}\text{'}\left(x\right)+P\left(x\right)P\text{'}\text{'}\text{'}\left(x\right)-P\text{'}\left(x\right)P\text{'}\text{'}\left(x\right)\\ & =& P\left(x\right)P\text{'}\text{'}\text{'}\left(x\right)\end{array}$

So the value of $2\frac{d}{dx}\left(y^3\frac{d^{2}y}{d{x}^2}\right)$ is $\begin{array}{rcl}& & P\left(x\right)P\text{'}\text{'}\text{'}\left(x\right)\end{array}$.

Hence, the correct option is C.