Question

2 sec x + 3 cot x − 4 tan x

Answer 1

$$\begin{gathered} \frac{d}{dx}\left(2secx+3cotx-4\tan x\right) \\ =2\frac{d}{dx}\left(\sec x\right)+3\frac{d}{dx}\left(\cot x\right)-4\frac{d}{dx}\left(\tan x\right) \\ =2\sec x\tan x-3\cos e{c}^{2}x-4{\sec }^{2}x \end{gathered}$$