If f(x) = | cases 2 ^ 4 x & 2 ^ 4 x & 3 + ^ 2 2x 3 + 2 ^ 4 x & 2 ^...

Given: f\left(x\right) = \left|\begin{cases}2\cos^{4}x & 2\sin^{4}x & 3 + \sin^{2}2x \\ 3 + 2\cos^{4}x & 2\sin^{4}x & \sin^{2}2x \\ 2\cos^{4}x & 3 + 2\sin^{4}x & \sin^{2}2x\end{cases}\right|

Applying ${R}_{2} \to {R}_{2} - {R}_{1}, {R}_{3} \to {R}_{3} - {R}_{1}$

⇒ f\left(x\right) = \left|\begin{cases}2\cos^{4}x & 2\sin^{4}x & 3 + \sin^{2}2x \\ 3 & 0 & - 3 \\ 0 & 3 & - 3\end{cases}\right|

⇒ f\left(x\right) = 9\left|\begin{cases}2\cos^{4}x & 2\sin^{4}x & 3 + \sin^{2}2x \\ 1 & 0 & - 1 \\ 0 & 1 & - 1\end{cases}\right|

$⇒ f\left(x\right) = 9\left(2\cos^{4}x + 2\sin^{4}x + 3 + \sin^{2}2x\right)$

Now, differentiating above function we get,

$⇒ {f}^{'}\left(x\right) = 9\left[8\cos^{3}x\left(- \sin x\right) + 8\sin^{3}x\left(\cos x\right) + 2\sin 2x\left(\cos 2x\right)\left(2\right)\right]$

$⇒ {f}^{'}\left(0\right) = 9\left[0 + 0 + 0\right]$

$⇒ {f}^{'}\left(0\right) = 0$