Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}u=x\tan x;v=\sec x+\tan x \\ \mathrm{Then},u\text{'}=x{\sec }^{2}x+\tan x;v\text{'}=\sec x\tan x+{\sec }^{2}x \\ \text{Using the quotient rule:} \\ \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu\text{'}-uv\text{'}}{v^2} \\ \frac{d}{dx}\left(\frac{x\tan x}{\sec x+\tan x}\right)=\frac{\left(\sec x+\tan x\right)\left(x{\sec }^{2}x+\tan x\right)-x\tan x\left(\sec x\tan x+{\sec }^{2}x\right)}{{\left(\sec x+\tan x\right)}^2} \\ =\frac{x{\sec }^{3}x+x{\sec }^{2}x\tan x+\sec x\tan x+{\tan }^{2}x-x\sec x{\tan }^{2}x-x\tan x{\sec }^{2}x}{{\left(\sec x+\tan x\right)}^2} \\ =\frac{\left(\sec x+\tan x\right)\left(x{\sec }^{2}x+\tan x\right)-x\tan x\sec x\left(\sec x+\tan x\right)}{{\left(\sec x+\tan x\right)}^2} \\ =\frac{x{\sec }^{2}x+\tan x-x\tan x\sec x}{\sec x+\tan x} \\ =\frac{x\sec x\left(\sec x-\tan x\right)+\tan x}{\sec x+\tan x} \end{gathered}$$