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Question
(1 − 2 tan x) (5 + 4 sin x)
Solution
\[\text{ Let } u = 1 - 2 \tan x; v = 5 + 4 \sin x \]
\[\text{ Then }, u' = - 2 \sec^2 x; v' = 4 \cos x\]
\[\text{ Using the product rule }:\]
\[\frac{d}{dx}\left( uv \right) = u v' + v u'\]
\[\frac{d}{dx}\left[ \left( 1 - 2 \tan x \right)\left( 5 + 4 \sin x \right) \right]\]
\[ = \left( 1 - 2 \tan x \right)\left( 4 \cos x \right) + \left( 5 + 4 \sin x \right)\left( - 2 \sec^2 x \right)\]
\[ = 4 \cos x - 8 \times \frac{\sin x}{\cos x}\cos x - 10 \sec^2 x - 8 \times \frac{\sin x}{\cos^2 x}\]
\[ = 4 \cos x - 8 \sin x - 10 \sec^2 x - 8 \sec x \tan x \]
\[ = 4\left( \cos x - 2 \sin x - \frac{5}{2} \sec^2 x - 2 \sec x \tan x \right)\]
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