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If $\mathrm{y}^{2}=\mathrm{p}(\mathrm{x})$ is a polynomial of degree 3, then what is
$2 \frac{\mathrm{d}}{\mathrm{dx}}\left[\mathrm{y}^{3} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right]$ equal to ?
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$2 \frac{\mathrm{d}}{\mathrm{dx}}\left[\mathrm{y}^{3} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right]$ equal to ?
Solution:
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Verified Answer
The correct answer is:
$\mathrm{p}(\mathrm{x}) \mathrm{p}^{\prime \prime \prime}(\mathrm{x})$
Given that $y^{2}=p(x)$
Differentiating
$\Rightarrow 2 \mathrm{yy}_{1}=\mathrm{p}^{\prime}(\mathrm{x})$
$\left[\right.$ here $\left.\mathrm{y}_{1}=\frac{\mathrm{dy}}{\mathrm{dx}}\right]$
$\Rightarrow \quad 2 \mathrm{y}_{1}=\frac{\mathrm{p}^{\prime}(\mathrm{x})}{\mathrm{y}}$
Differentiating again, $\Rightarrow \quad 2 \mathrm{y}_{2}=\frac{\mathrm{yp}^{\prime \prime}(\mathrm{x})-\mathrm{p}^{\prime}(\mathrm{x}) \mathrm{y}_{1}}{\mathrm{y}^{2}},\left[\mathrm{y}_{2}=\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}}\right]$
$\Rightarrow 2 \mathrm{y}_{2}=\frac{\mathrm{yp}^{\prime \prime}(\mathrm{x})-\frac{\mathrm{p}^{\prime}(\mathrm{x}) \cdot \mathrm{p}^{\prime}(\mathrm{x})}{2 \mathrm{y}}}{\mathrm{y}^{2}}$
$$
\begin{array}{l}
\quad=\frac{\left.2 \mathrm{y}^{2} \mathrm{p}^{\prime \prime}(\mathrm{x})-\mathrm{p}^{\prime}(\mathrm{x})\right)^{2}}{2 \mathrm{y}^{3}} \\
\Rightarrow \quad 2 \mathrm{y}^{3} \mathrm{y}_{2}=\frac{1}{2}\left[2 \mathrm{y}^{2} \mathrm{p}^{\prime \prime}(\mathrm{x})-\left(\mathrm{p}^{\prime}(\mathrm{x})\right)^{2}\right] \\
\Rightarrow \quad 2 \mathrm{y}^{3} \mathrm{y}_{2}=\frac{1}{2}\left[2 \mathrm{p}(\mathrm{x}) \mathrm{p}^{\prime \prime}(\mathrm{x})-\left(\mathrm{p}^{\prime}(\mathrm{x})\right)^{2}\right] \\
\Rightarrow \quad 2 \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{y}^{3} \mathrm{y}_{2}\right) \\
=\frac{1}{2}\left[2 \mathrm{p}^{\prime}(\mathrm{x}) \mathrm{p}^{\prime \prime}(\mathrm{x})+2 \mathrm{p}(\mathrm{x}) \mathrm{p}^{\prime \prime}(\mathrm{x})-2 \mathrm{p}^{\prime}(\mathrm{x}) \mathrm{p}^{\prime \prime}(\mathrm{x})\right] \\
=\mathrm{p}(\mathrm{x}) \mathrm{p}^{\prime \prime \prime}(\mathrm{x})
\end{array}
$$
Differentiating
$\Rightarrow 2 \mathrm{yy}_{1}=\mathrm{p}^{\prime}(\mathrm{x})$
$\left[\right.$ here $\left.\mathrm{y}_{1}=\frac{\mathrm{dy}}{\mathrm{dx}}\right]$
$\Rightarrow \quad 2 \mathrm{y}_{1}=\frac{\mathrm{p}^{\prime}(\mathrm{x})}{\mathrm{y}}$
Differentiating again, $\Rightarrow \quad 2 \mathrm{y}_{2}=\frac{\mathrm{yp}^{\prime \prime}(\mathrm{x})-\mathrm{p}^{\prime}(\mathrm{x}) \mathrm{y}_{1}}{\mathrm{y}^{2}},\left[\mathrm{y}_{2}=\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}}\right]$
$\Rightarrow 2 \mathrm{y}_{2}=\frac{\mathrm{yp}^{\prime \prime}(\mathrm{x})-\frac{\mathrm{p}^{\prime}(\mathrm{x}) \cdot \mathrm{p}^{\prime}(\mathrm{x})}{2 \mathrm{y}}}{\mathrm{y}^{2}}$
$$
\begin{array}{l}
\quad=\frac{\left.2 \mathrm{y}^{2} \mathrm{p}^{\prime \prime}(\mathrm{x})-\mathrm{p}^{\prime}(\mathrm{x})\right)^{2}}{2 \mathrm{y}^{3}} \\
\Rightarrow \quad 2 \mathrm{y}^{3} \mathrm{y}_{2}=\frac{1}{2}\left[2 \mathrm{y}^{2} \mathrm{p}^{\prime \prime}(\mathrm{x})-\left(\mathrm{p}^{\prime}(\mathrm{x})\right)^{2}\right] \\
\Rightarrow \quad 2 \mathrm{y}^{3} \mathrm{y}_{2}=\frac{1}{2}\left[2 \mathrm{p}(\mathrm{x}) \mathrm{p}^{\prime \prime}(\mathrm{x})-\left(\mathrm{p}^{\prime}(\mathrm{x})\right)^{2}\right] \\
\Rightarrow \quad 2 \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{y}^{3} \mathrm{y}_{2}\right) \\
=\frac{1}{2}\left[2 \mathrm{p}^{\prime}(\mathrm{x}) \mathrm{p}^{\prime \prime}(\mathrm{x})+2 \mathrm{p}(\mathrm{x}) \mathrm{p}^{\prime \prime}(\mathrm{x})-2 \mathrm{p}^{\prime}(\mathrm{x}) \mathrm{p}^{\prime \prime}(\mathrm{x})\right] \\
=\mathrm{p}(\mathrm{x}) \mathrm{p}^{\prime \prime \prime}(\mathrm{x})
\end{array}
$$
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