Question

$\frac{x+\cos x}{\tan x}$

Answer 1

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