Question

Differentiate $\log (\sec x+\tan x)$ with respect to $x$

Answer 1

Using chain rule,

$\frac{d\left[\log \right(\sec x+\tan x\left)\right]}{dx}=\frac{1}{\sec x+\tan x}\times \frac{d(\sec x+\tan x)}{dx}$

$=\frac{\sec x\tan x+{\sec }^{2}x}{\sec x+\tan x}$

$=\sec x$