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Derivatives

Question

$If f (x) = |\begin{array}{ccc} \sec^{2} x & 1 & 1 \\ \cos^{2} x & \cos^{2} x & \cos e c^{2} x \\ 1 & \cos^{2} x & \cot^{2} x \end{array}| then f^{'} (x) =$

Moderate

A

$\cos^{4} x$
B

$- 4 \cos^{3} x \sin x$
C

$4 \cos^{3} x \sin x$
D

$4 \sin^{3} x \cos x$

Solution

$f (x) = |\begin{matrix} \sec^{2} x & 1 & 1 \\ \cos^{2} x & \cos^{2} x & \cos e c^{2} x \\ \begin{matrix} x \\ 1 \end{matrix} & \cos^{2} x & \cot^{2} x \end{matrix}|$
$\begin{array}{l} = \sec^{2} x [\cos^{2} x \cdot \cot^{2} x - \cos^{2} x \cos e c^{2} x] - 1 [\cos^{2} x \cdot \cot^{2} x - \cos e c^{2} x] + 1 [\cos^{4} x - \cos^{2} x] \\ = \cot^{2} x - \cos e c^{2} x - \cos^{2} x \cot^{2} x + \cos e c^{2} x + \cos^{4} x - \cos^{2} x = \cot^{2} x (1 - \cos^{2} x) + \cos^{4} x - \cos^{2} x \\ = \cot^{2} x - \cos e c^{2} x - \cos^{2} x \cot^{2} x + \cos e c^{2} x + \cos^{4} x - \cos^{2} x = \cot^{2} x (1 - \cos^{2} x) + \cos^{4} x - \cos^{2} x \\ = \cos^{2} x + \cos^{4} x - \cos^{2} x \\ = \cos^{4} x . \end{array}$
$f' (x) = - 4 \cos^{3} x \sin x$

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