Question

If ${{y}^{2}}=P\left( x \right)$ is a polynomial of degree 3, then $2\left( \dfrac{d}{dx} \right)\left( {{y}^{3}}\centerdot \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)$is equal to:
(a) ${P}'''\left( x \right)+{P}'\left( x \right)$
(b) ${P}''\left( x \right)\centerdot {P}'''\left( x \right)$
(c) $P\left( x \right)\centerdot {P}'''\left( x \right)$
(d) A constant

Answer 1

Hint:First, we before proceeding for this, we must find the first derivative of the given polynomial which is ${{y}^{2}}=P\left( x \right)$with respect to x. Then, again by using the product rule $\dfrac{d}{dx}\left( uv \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ to get the second derivative of the above function, we get the value of second derivate. Then, by substituting the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$in the given question and then differentiating, we get the desired value.

Complete step by step answer:
In this question, we are supposed to find the value of expression $2\left( \dfrac{d}{dx} \right)\left( {{y}^{3}}\centerdot \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)$when ${{y}^{2}}=P\left( x \right)$ is a polynomial of degree 3.
So, before proceeding for this, we must find the first derivative of the given polynomial which is ${{y}^{2}}=P\left( x \right)$with respect to x as:
$2y\dfrac{dy}{dx}={P}'\left( x \right)$
Then, again by using the product rule $\dfrac{d}{dx}\left( uv \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ to get the second derivative of the above function as:
$2y\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2{{\left( \dfrac{dy}{dx} \right)}^{2}}={P}''\left( x \right)$
Now, we need to find the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ from the above expression, we get:
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{{P}''\left( x \right)-2{{\left( \dfrac{dy}{dx} \right)}^{2}}}{2y}$
Then, in the question we need to find the value of the equation $\left( {{y}^{3}}\centerdot \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)$.
So, by substituting the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$found above, we get:
$\begin{align}
  & \left( {{y}^{3}}\centerdot \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)=\dfrac{{{y}^{3}}{P}''\left( x \right)}{2y}-\dfrac{{{y}^{3}}{{\left[ {P}'\left( x \right) \right]}^{2}}}{4{{y}^{3}}} \\
 & \Rightarrow \left( {{y}^{3}}\centerdot \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)=\dfrac{{{y}^{2}}{P}''\left( x \right)}{2}-\dfrac{{{\left[ {P}'\left( x \right) \right]}^{2}}}{4} \\
\end{align}$
Now, in the question we need to differentiate the $\left( \dfrac{d}{dx} \right)\left( {{y}^{3}}\centerdot \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)$above obtained value as:
$\dfrac{d}{dx}\left( {{y}^{3}}\centerdot \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)=\dfrac{d}{dx}\left( \dfrac{{{y}^{2}}{P}''\left( x \right)}{2}-\dfrac{{{\left[ {P}'\left( x \right) \right]}^{2}}}{4} \right)$
Then, again by applying the product rule and replacing ${{y}^{2}}=P\left( x \right)$, we get:
$\begin{align}
  & \dfrac{d}{dx}\left( {{y}^{3}}\centerdot \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)=\dfrac{d}{dx}\left( \dfrac{P\left( x \right)\centerdot {P}''\left( x \right)}{2}-\dfrac{{{\left[ {P}'\left( x \right) \right]}^{2}}}{4} \right) \\
 & \Rightarrow \dfrac{d}{dx}\left( {{y}^{3}}\centerdot \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)=\dfrac{{P}'\left( x \right)\centerdot {P}''\left( x \right)+P\left( x \right)\centerdot {P}'''\left( x \right)}{2}-\dfrac{2{P}'\left( x \right)\centerdot {P}''\left( x \right)}{4} \\
\end{align}$
Now, we need to find the value of $2\left( \dfrac{d}{dx} \right)\left( {{y}^{3}}\centerdot \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)$, so multiplying by 2 in above expression, we get:
$\begin{align}
  & 2\dfrac{d}{dx}\left( {{y}^{3}}\centerdot \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)=\dfrac{2{P}'\left( x \right)\centerdot {P}''\left( x \right)+P\left( x \right)\centerdot {P}'''\left( x \right)}{2}-\dfrac{4{P}'\left( x \right)\centerdot {P}''\left( x \right)}{4} \\
 & \Rightarrow 2\dfrac{d}{dx}\left( {{y}^{3}}\centerdot \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)={P}'\left( x \right)\centerdot {P}''\left( x \right)+P\left( x \right)\centerdot {P}'''\left( x \right)-{P}'\left( x \right)\centerdot {P}''\left( x \right) \\
 & \Rightarrow 2\dfrac{d}{dx}\left( {{y}^{3}}\centerdot \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)=P\left( x \right)\centerdot {P}'''\left( x \right) \\
\end{align}$
So, we get the value of$2\left( \dfrac{d}{dx} \right)\left( {{y}^{3}}\centerdot \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)$ as $P\left( x \right)\centerdot {P}'''\left( x \right)$.
Hence, the option (c) is correct.

Note: Now, to solve these type of the questions we need to know some of the basics of differentiation beforehand so that question can be solved easily. So, the basics required are as:
$\begin{align}
  & \dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}} \\
 & \dfrac{d}{dx}f\left( x \right)={f}'\left( x \right) \\
\end{align}$