Question

Find the derivative of $\frac{\sec x-1}{\sec x+1}$.

Answer 1

Let $f\left(x\right)=\frac{\sec x-1}{\sec x+1}$
Differentiating with respect to $x$
$\Rightarrow f^{\prime }\left(x\right)=\frac{d}{dx}\left(\frac{\sec x-1}{\sec x+1}\right)$
$\Rightarrow f^{\prime }\left(x\right)=\frac{(\sec x+1)\frac{d}{dx}(\sec x-1)-(\sec x-1)\frac{d}{dx}(\sec x+1)}{(\sec x+1{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{\left[\right(\sec x+1\left)\right(\sec x\tan x)-(\sec x-1\left)\right(\sec x\tan x\left)\right]}{(\sec x+1{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{\sec x\tan x\left[\right(\sec x+1)-(\sec x-1\left)\right]}{(\sec x+1{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{2\sec x\tan x}{(\sec x+1{)}^2}$
$\therefore f^{\prime }\left(x\right)=\frac{2\sec x\tan x}{(\sec x+1{)}^2}$