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If the functions $p\left( x \right),q\left( x \right),r\left( x \right)$ are three polynomials of degree 2, then prove that
$\left| \begin{matrix}
   p\left( x \right) & q\left( x \right) & r\left( x \right) \\
   p'\left( x \right) & q'\left( x \right) & r'\left( x \right) \\
   p''\left( x \right) & q''\left( x \right) & r''\left( x \right) \\
\end{matrix} \right|$ is independent of x.

Answer Verified Verified
Hint: To solve this question, we should the way to differentiate a determinant. Let us consider the whole determinant as $f\left( x \right)$. For a function $v\left( x \right)$ such that
$v\left( x \right)=\left| \begin{matrix}
   p\left( x \right) & q\left( x \right) & r\left( x \right) \\
   s\left( x \right) & t\left( x \right) & u\left( x \right) \\
   w\left( x \right) & x\left( x \right) & y\left( x \right) \\
\end{matrix} \right|$
$v'\left( x \right)$ is defined as
$v'\left( x \right)=\left| \begin{matrix}
   p'\left( x \right) & q'\left( x \right) & r'\left( x \right) \\
   s\left( x \right) & t\left( x \right) & u\left( x \right) \\
   w\left( x \right) & x\left( x \right) & y\left( x \right) \\
\end{matrix} \right|+\left| \begin{matrix}
   p\left( x \right) & q\left( x \right) & r\left( x \right) \\
   s'\left( x \right) & t'\left( x \right) & u'\left( x \right) \\
   w\left( x \right) & x\left( x \right) & y\left( x \right) \\
\end{matrix} \right|+\left| \begin{matrix}
   p\left( x \right) & q\left( x \right) & r\left( x \right) \\
   s\left( x \right) & t\left( x \right) & u\left( x \right) \\
   w'\left( x \right) & x'\left( x \right) & y'\left( x \right) \\
\end{matrix} \right|$
Using this formula on the equation
$f\left( x \right)=\left| \begin{matrix}
   p\left( x \right) & q\left( x \right) & r\left( x \right) \\
   p'\left( x \right) & q'\left( x \right) & r'\left( x \right) \\
   p''\left( x \right) & q''\left( x \right) & r''\left( x \right) \\
\end{matrix} \right|$
and the property that the third derivative of a second order polynomial is zero, we get that $f'\left( x \right)=0$.
This means that f(x) is a constant function and it doesn’t depend on x.

Complete step-by-step answer:
Let us consider a function $v\left( x \right)$ such that,
$v\left( x \right)=\left| \begin{matrix}
   p\left( x \right) & q\left( x \right) & r\left( x \right) \\
   s\left( x \right) & t\left( x \right) & u\left( x \right) \\
   w\left( x \right) & x\left( x \right) & y\left( x \right) \\
\end{matrix} \right|$
Let us consider the derivative of $v\left( x \right)$. $v'\left( x \right)$ is defined as
$v'\left( x \right)=\left| \begin{matrix}
   p'\left( x \right) & q'\left( x \right) & r'\left( x \right) \\
   s\left( x \right) & t\left( x \right) & u\left( x \right) \\
   w\left( x \right) & x\left( x \right) & y\left( x \right) \\
\end{matrix} \right|+\left| \begin{matrix}
   p\left( x \right) & q\left( x \right) & r\left( x \right) \\
   s'\left( x \right) & t'\left( x \right) & u'\left( x \right) \\
   w\left( x \right) & x\left( x \right) & y\left( x \right) \\
\end{matrix} \right|+\left| \begin{matrix}
   p\left( x \right) & q\left( x \right) & r\left( x \right) \\
   s\left( x \right) & t\left( x \right) & u\left( x \right) \\
   w'\left( x \right) & x'\left( x \right) & y'\left( x \right) \\
\end{matrix} \right|\to \left( 1 \right)$

Let us consider the determinant given in the question as f(x).
$f\left( x \right)=\left| \begin{matrix}
   p\left( x \right) & q\left( x \right) & r\left( x \right) \\
   p'\left( x \right) & q'\left( x \right) & r'\left( x \right) \\
   p''\left( x \right) & q''\left( x \right) & r''\left( x \right) \\
\end{matrix} \right|$
Let us consider differentiating the function f(x), from equation-1, we get
$f'\left( x \right)=\left| \begin{matrix}
   p'\left( x \right) & q'\left( x \right) & r'\left( x \right) \\
   p'\left( x \right) & q'\left( x \right) & r'\left( x \right) \\
   p''\left( x \right) & q''\left( x \right) & r''\left( x \right) \\
\end{matrix} \right|+\left| \begin{matrix}
   p\left( x \right) & q\left( x \right) & r\left( x \right) \\
   p''\left( x \right) & q''\left( x \right) & r''\left( x \right) \\
   p''\left( x \right) & q''\left( x \right) & r''\left( x \right) \\
\end{matrix} \right|+\left| \begin{matrix}
   p\left( x \right) & q\left( x \right) & r\left( x \right) \\
   p'\left( x \right) & q'\left( x \right) & r'\left( x \right) \\
   p'''\left( x \right) & q'''\left( x \right) & r'''\left( x \right) \\
\end{matrix} \right|\to \left( 2 \right)$
We know the property of determinants that the value of determinant having any two rows or columns equal is zero.
$v\left( x \right)=\left| \begin{matrix}
   p\left( x \right) & q\left( x \right) & r\left( x \right) \\
   p\left( x \right) & q\left( x \right) & r\left( x \right) \\
   w\left( x \right) & x\left( x \right) & y\left( x \right) \\
\end{matrix} \right|=0$
Using this property in equation-2, we get
In the determinant $\left| \begin{matrix}
   p'\left( x \right) & q'\left( x \right) & r'\left( x \right) \\
   p'\left( x \right) & q'\left( x \right) & r'\left( x \right) \\
   p''\left( x \right) & q''\left( x \right) & r''\left( x \right) \\
\end{matrix} \right|$, the corresponding terms of the first and second rows are equal. Similarly, in the determinant $\left| \begin{matrix}
   p\left( x \right) & q\left( x \right) & r\left( x \right) \\
   p''\left( x \right) & q''\left( x \right) & r''\left( x \right) \\
   p''\left( x \right) & q''\left( x \right) & r''\left( x \right) \\
\end{matrix} \right|$, the terms of the second and third row are equal.

$\begin{align}
  & f'\left( x \right)=0+0+\left| \begin{matrix}
   p\left( x \right) & q\left( x \right) & r\left( x \right) \\
   p'\left( x \right) & q'\left( x \right) & r'\left( x \right) \\
   p'''\left( x \right) & q'''\left( x \right) & r'''\left( x \right) \\
\end{matrix} \right| \\
 & f'\left( x \right)=\left| \begin{matrix}
   p\left( x \right) & q\left( x \right) & r\left( x \right) \\
   p'\left( x \right) & q'\left( x \right) & r'\left( x \right) \\
   p'''\left( x \right) & q'''\left( x \right) & r'''\left( x \right) \\
\end{matrix} \right|\to \left( 3 \right) \\
\end{align}$
Let us consider a second order polynomial in x
$h(x)=a{{x}^{2}}+bx+c$.
Differentiate $h\left( x \right)$ until its third derivative.
$\begin{align}
  & h(x)=a{{x}^{2}}+bx+c \\
 & h'(x)=2ax+b \\
 & h''\left( x \right)=2a \\
 & h'''\left( x \right)=0 \\
\end{align}$
This tells us that the third derivative of any second order polynomial in x is zero.
Using this relation in equation-3, p(x), q(x), r(x) are second degree polynomials, we get
$f'\left( x \right)=\left| \begin{matrix}
   p\left( x \right) & q\left( x \right) & r\left( x \right) \\
   p'\left( x \right) & q'\left( x \right) & r'\left( x \right) \\
   0 & 0 & 0 \\
\end{matrix} \right|$
We know that in a determinant, if any of the rows or columns has all zeros, the determinant value is also zero.
So, we get
$f'(x)=0$
Integrating on both sides with respect to x, we get
$\begin{align}
  & \int{f'(x)}=\int{0} \\
 & f\left( x \right)=c \\
\end{align}$
We got the value of f(x) as constant which means that the function f(x) is independent of x.
$\therefore $Hence proved the statement that f(x) is independent of x.

Note: Some students tend to expand the determinant and then try to differentiate the function which leads to a confusion. Instead using the above mentioned differentiation property reduces the complexity. After getting the first two determinants to zero, some students cannot proceed from $f'\left( x \right)=\left| \begin{matrix}
   p\left( x \right) & q\left( x \right) & r\left( x \right) \\
   p'\left( x \right) & q'\left( x \right) & r'\left( x \right) \\
   p'''\left( x \right) & q'''\left( x \right) & r'''\left( x \right) \\
\end{matrix} \right|$ as they overlook that all the functions p(x), q(x), r(x) are of second degree. So, each and every word in the question is important.