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

& \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{aligned}$$ Now , $$\begin{aligned} & \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{aligned}$$ Now , we will determine the value of$$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{aligned} & {{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{aligned}$$ $$=\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{aligned} & f'(0)=\\{-1(3-2)\\}+\\{1(6-0)+1(0-2)\\}+\\{1(0-6)+1(3-0)\\} \\\ & =-1+6-2-6+3 \\\ & =0 \\\ \end{aligned}$$ Now we will determine the value of$$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{aligned} & {{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{aligned}$$ $$\begin{aligned} & =\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{aligned}$$ $$=\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 .