Question

Differentiate with respective to $x$:

$\log (\sec x+\tan x)$

Answer 1

Let $y=\log (\sec x+\tan x)$

we know that $\frac{d\left(\sec x\right)}{dx}=\sec x\tan x,\frac{d\left(\tan x\right)}{dx}={\sec }^{2}x,\frac{d\left(\log x\right)}{dx}=\frac{1}{x}$

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

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