Question

If $y = \log(\sec x + \tan x),$ then $\frac{dy}{dx} =$

• A. $\tan x$
• B. $sec x$
• C. $\tan^2 x$
• D. none of these

Answer 1

Answer: (B)

$sec x$

Answer 2

$\frac{dy}{dx} = \frac{1}{\sec x + \tan x} . \left(\sec x \tan x + \sec^{2} x\right)$ $= \frac{\sec x\left[\tan x +\sec x\right]}{\sec x +\tan x} = \sec x$