Question

Differentiate the following w.r.t.x:
${\tan }^{-1}(\sec x+\tan x)$

Answer 1

$y={\tan }^{-1}(\sec x+\tan x)$

Differentiating w.r.t. x,

$\frac{dy}{dx}=\frac{1}{1+(\sec x+\tan x{)}^2}(\sec x\tan x+{\sec }^{2}x)$

$\frac{dy}{dx}=\frac{\sec x\tan x+{\sec }^{2}x}{1+(\sec x+\tan x{)}^2}$