Question

(1 − 2 tan x) (5 + 4 sin x)

Answer 1

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