1. The problem is to find the differential coefficient (derivative) of the function $y = \ln(\sec x + \tan x)$.\n\n2. Recall the chain rule for differentiation: if $y = \ln u$, then $\frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx}$.\n\n3. Let $u = \sec x + \tan x$. We need to find $\frac{du}{dx}$.\n\n4. The derivatives are $\frac{d}{dx}(\sec x) = \sec x \tan x$ and $\frac{d}{dx}(\tan x) = \sec^2 x$.\n\n5. Therefore, $\frac{du}{dx} = \sec x \tan x + \sec^2 x$.\n\n6. Applying the chain rule,\n$$\frac{dy}{dx} = \frac{1}{\sec x + \tan x} \cdot (\sec x \tan x + \sec^2 x)$$\n\n7. Factor the numerator:\n$$\sec x \tan x + \sec^2 x = \sec x (\tan x + \sec x)$$\n\n8. Substitute back:\n$$\frac{dy}{dx} = \frac{\sec x (\tan x + \sec x)}{\sec x + \tan x}$$\n\n9. Notice $\tan x + \sec x = \sec x + \tan x$, so numerator and denominator are the same except for the factor $\sec x$.\n\n10. Cancel the common factor $\sec x + \tan x$:\n$$\frac{dy}{dx} = \cancel{\frac{\sec x (\tan x + \sec x)}{\sec x + \tan x}} = \sec x$$\n\n11. Final answer: $$\frac{dy}{dx} = \sec x$$