Question

Let a function is given as \[f(x)=\left| \begin{matrix}
   \cos x & \sin x & \cos x \\
   \cos 2x & \sin 2x & 2\cos 2x \\
   \cos 3x & \sin 3x & 3\cos 3x \\
\end{matrix} \right|\], then find the value of \[f'(0)\] and \[f'\left( \dfrac{\pi }{2} \right)\].

Answer 1

Hint: If we write a determinant as \[\Delta =\left| \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|\], then the derivative of the determinant can be written as \[\dfrac{d}{dx}\left( \Delta \right)=\left| \begin{matrix}
   a_{1}^{'} & a_{2}^{'} & a_{3}^{'} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|+\left| \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   b_{1}^{'} & b_{2}^{'} & b_{3}^{'} \\
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|+\left| \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   c_{1}^{'} & c_{2}^{'} & c_{3}^{'} \\
\end{matrix} \right|\].

Complete step by step answer:
We are given \[f(x)=\left| \begin{matrix}
   \cos x & \sin x & \cos x \\
   \cos 2x & \sin 2x & 2\cos 2x \\
   \cos 3x & \sin 3x & 3\cos 3x \\
\end{matrix} \right|\]. We need to find the value of \[f'(0)\] and \[f'\left( \dfrac{\pi }{2} \right)\].
To find the value of \[f'(0)\] and \[f'\left( \dfrac{\pi }{2} \right)\] , first , we need to determine the value of \[f'(x)\].
Now , to find the value of \[f'(x)\], we will differentiate the given function with respect to \[x\] .
We know , if \[\Delta =\left| \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|\], then \[\dfrac{d}{dx}\left( \Delta \right)=\left| \begin{matrix}
   a_{1}^{'} & a_{2}^{'} & a_{3}^{'} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|+\left| \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   b_{1}^{'} & b_{2}^{'} & b_{3}^{'} \\
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|+\left| \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   c_{1}^{'} & c_{2}^{'} & c_{3}^{'} \\
\end{matrix} \right|\]
So , on differentiating the given determinant with respect to \[x\], we get
\[\begin{align}
  & \dfrac{d}{dx}f(x)=\left| \begin{matrix}
   \dfrac{d}{dx}\left( \cos x \right) & \dfrac{d}{dx}\left( \sin x \right) & \dfrac{d}{dx}\left( \cos x \right) \\
   \cos 2x & \sin 2x & 2\cos 2x \\
   \cos 3x & \sin 3x & 3\cos 3x \\
\end{matrix} \right|+\left| \begin{matrix}
   \cos x & \sin x & \cos x \\
   \dfrac{d}{dx}\left( \cos 2x \right) & \dfrac{d}{dx}\left( \sin 2x \right) & \dfrac{d}{dx}\left( 2\cos 2x \right) \\
   \cos 3x & \sin 3x & 3\cos 3x \\
\end{matrix} \right|+ \\
 & \left| \begin{matrix}
   \cos x & \sin x & \cos x \\
   \cos 2x & \sin 2x & 2\cos 2x \\
   \dfrac{d}{dx}\left( \cos 3x \right) & \dfrac{d}{dx}\left( \sin 3x \right) & \dfrac{d}{dx}\left( 3\cos 3x \right) \\
\end{matrix} \right| \\
\end{align}\]
Now ,
\[\begin{align}
  & \dfrac{d}{dx}f(x)=\left| \begin{matrix}
   -\sin x & \cos x & -\sin x \\
   \cos 2x & \sin 2x & 2\cos 2x \\
   \cos 3x & \sin 3x & 3\cos 3x \\
\end{matrix} \right|+\left| \begin{matrix}
   \cos x & \sin x & \cos x \\
   -2\sin 2x & 2\cos 2x & -4\sin 2x \\
   \cos 3x & \sin 3x & 3\cos 3x \\
\end{matrix} \right|+ \\
 & \left| \begin{matrix}
   \cos x & \sin x & \cos x \\
   \cos 2x & \sin 2x & 2\cos 2x \\
   -3\sin 3x & 3\cos 3x & -9\sin 3x \\
\end{matrix} \right| \\
\end{align}\]
Now , we will determine the value of\[f'(0)\].
To evaluate\[f'(0)\], we will substitute \[0\]in place of \[x\] in the expression of \[f'(x)\].
On substituting \[0\]in place of \[x\] in the expression of \[f'(x)\], we get
\[\begin{align}
  & {{f}^{'}}(0)=\left| \begin{matrix}
   -\sin 0 & \cos 0 & -\sin 0 \\
   \cos 2(0) & \sin 2(0) & 2\cos 2(0) \\
   \cos 3(0) & \sin 3(0) & 3\cos 3(0) \\
\end{matrix} \right|+\left| \begin{matrix}
   \cos 0 & \sin 0 & \cos 0 \\
   -2\sin 2(0) & 2\cos 2(0) & -4\sin 2(0) \\
   \cos 3(0) & \sin 3(0) & 3\cos 3(0) \\
\end{matrix} \right|+ \\
 & \left| \begin{matrix}
   \cos 0 & \sin 0 & \cos 0 \\
   \cos 2(0) & \sin 2(0) & 2\cos 2(0) \\
   -3\sin 3(0) & 3\cos 3(0) & -9\sin 3(0) \\
\end{matrix} \right| \\
\end{align}\]
\[=\left| \begin{matrix}
   0 & 1 & 0 \\
   1 & 0 & 2 \\
   1 & 0 & 3 \\
\end{matrix} \right|+\left| \begin{matrix}
   1 & 0 & 1 \\
   0 & 2 & 0 \\
   1 & 0 & 3 \\
\end{matrix} \right|+\left| \begin{matrix}
   1 & 0 & 1 \\
   1 & 0 & 2 \\
   0 & 3 & 0 \\
\end{matrix} \right|\]
Now , we will determine the values of the determinants , add them to determine the value of \[f'(0)\].
\[\begin{align}
  & f'(0)=\{-1(3-2)\}+\{1(6-0)+1(0-2)\}+\{1(0-6)+1(3-0)\} \\
 & =-1+6-2-6+3 \\
 & =0 \\
\end{align}\]
Now we will determine the value of\[f'\left( \dfrac{\pi }{2} \right)\].
To evaluate \[f'\left( \dfrac{\pi }{2} \right)\] , we will substitute \[\dfrac{\pi }{2}\]in place of \[x\] in the expression of \[f'(x)\].
On substituting \[\dfrac{\pi }{2}\]in place of \[x\] in the expression of \[f'(x)\], we get \[\begin{align}
  & {{f}^{'}}\left( \dfrac{\pi }{2} \right)=\left| \begin{matrix}
   -\sin (\dfrac{\pi }{2}) & \cos (\dfrac{\pi }{2}) & -\sin (\dfrac{\pi }{2}) \\
   \cos 2(\dfrac{\pi }{2}) & \sin 2(\dfrac{\pi }{2}) & 2\cos 2(\dfrac{\pi }{2}) \\
   \cos 3(\dfrac{\pi }{2}) & \sin 3(\dfrac{\pi }{2}) & 3\cos 3(\dfrac{\pi }{2}) \\
\end{matrix} \right|+\left| \begin{matrix}
   \cos (\dfrac{\pi }{2}) & \sin (\dfrac{\pi }{2}) & \cos (\dfrac{\pi }{2}) \\
   -2\sin 2(\dfrac{\pi }{2}) & 2\cos 2(\dfrac{\pi }{2}) & -4\sin 2(\dfrac{\pi }{2}) \\
   \cos 3(\dfrac{\pi }{2}) & \sin 3(\dfrac{\pi }{2}) & 3\cos 3(\dfrac{\pi }{2}) \\
\end{matrix} \right|+ \\
 & \left| \begin{matrix}
   \cos (\dfrac{\pi }{2}) & \sin (\dfrac{\pi }{2}) & \cos (\dfrac{\pi }{2}) \\
   \cos 2(\dfrac{\pi }{2}) & \sin 2(\dfrac{\pi }{2}) & 2\cos 2(\dfrac{\pi }{2}) \\
   -3\sin 3(\dfrac{\pi }{2}) & 3\cos 3(\dfrac{\pi }{2}) & -9\sin 3(\dfrac{\pi }{2}) \\
\end{matrix} \right| \\
\end{align}\]
\[\begin{align}
  & =\left| \begin{matrix}
   -\sin (\dfrac{\pi }{2}) & \cos (\dfrac{\pi }{2}) & -\sin (\dfrac{\pi }{2}) \\
   \cos \pi & \sin \pi & 2\cos \pi \\
   \cos 3(\dfrac{\pi }{2}) & \sin 3(\dfrac{\pi }{2}) & 3\cos 3(\dfrac{\pi }{2}) \\
\end{matrix} \right|+\left| \begin{matrix}
   \cos (\dfrac{\pi }{2}) & \sin (\dfrac{\pi }{2}) & \cos (\dfrac{\pi }{2}) \\
   -2\sin \pi & 2\cos \pi & -4\sin \pi \\
   \cos 3(\dfrac{\pi }{2}) & \sin 3(\dfrac{\pi }{2}) & 3\cos 3(\dfrac{\pi }{2}) \\
\end{matrix} \right|+ \\
 & \left| \begin{matrix}
   \cos (\dfrac{\pi }{2}) & \sin (\dfrac{\pi }{2}) & \cos (\dfrac{\pi }{2}) \\
   \cos \pi & \sin \pi & 2\cos \pi \\
   -3\sin 3(\dfrac{\pi }{2}) & 3\cos 3(\dfrac{\pi }{2}) & -9\sin 3(\dfrac{\pi }{2}) \\
\end{matrix} \right| \\
\end{align}\]
\[=\left| \begin{matrix}
   -1 & 0 & -1 \\
   -1 & 0 & -2 \\
   0 & -1 & 0 \\
\end{matrix} \right|+\left| \begin{matrix}
   0 & 1 & 0 \\
   0 & -2 & 0 \\
   0 & -1 & 0 \\
\end{matrix} \right|+\left| \begin{matrix}
   0 & 1 & 0 \\
   -1 & 0 & -2 \\
   3 & 0 & 9 \\
\end{matrix} \right|\]
Now , we will determine the values of the determinants , add them to determine the value of \[{{f}^{'}}\left( \dfrac{\pi }{2} \right)\].
\[{{f}^{'}}\left( \dfrac{\pi }{2} \right)=\{-1(0-2)-0(0-0)-1(1-0\}+\{0(0-0)-1(0-0)+0(0-0)\}+\{0(0-0)-1(-9+6)+0(0-0)\}\]
\[=2-1+3\]
\[=4\]
Hence , the value of\[f'(0)\] is \[0\] and the value of\[f'(\dfrac{\pi }{2})\] is \[4\] .

Note: Always remember that the derivative of \[\sin x\]is \[\cos x\]and not \[-\cos x\]. Also, the derivative of \[\cos x\] is\[-\sin x\]and not \[\sin x\]. Students often get confused between the two and make mistakes . Due to such mistakes , students often end up getting a wrong answer. So , such mistakes should be avoided .