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

Solution

1. Given $y=\ln (\sec x+\tan x)$, apply the chain rule to find the derivative: $\frac{dy}{dx}=\frac{1}{\sec x+\tan x}\cdot \frac{d}{dx}(\sec x+\tan x)$.
2. Calculate the derivative of the inner function: $\frac{d}{dx}(\sec x+\tan x)=\sec x\tan x+{\sec }^{2}x$.
3. Factor out $\sec x$ from the numerator to get $\frac{\sec x(\tan x+\sec x)}{\sec x+\tan x}$.
4. Cancel the common term $(\sec x+\tan x)$ to simplify the expression to $\sec x$.
5. Hence the answer is (A).