In which interval \(\mathrm{f}(\mathrm{x})=\tan ^{-1}(\sin \mathrm{x}+\cos \mathrm{x})\) is strictly increasing? (a) \([\\{(-\pi) / 2\\}, 0]\) (b) \([0,(\pi / 4)]\) (c) \([\\{(-\pi) / 2\\},(\pi / 2)]\) (d) \((-\pi, \pi)\)

Understanding the first derivative test is pivotal in determining whether a function is increasing or decreasing at particular points. This is a common method used in calculus to assess the behavior of functions.

When we are given a function, like in our exercise \(f(x) = \tan^{-1}(\sin x + \cos x)\), the first derivative, denoted as \( f'(x) \), gives us the function's rate of change. A positive rate of change suggests that the function is increasing, while a negative one indicates a decrease. By finding where the derivative of a function is positive, we can pinpoint intervals where the function itself is on the rise.

We can summarize the first derivative test in three major steps:

• Find the first derivative of the function.
• Set the derivative greater than zero to determine where the function is increasing.
• Analyze the critical points and intervals to understand the function's behavior.

In the provided exercise, the correct interval that makes \(f'(x) > 0\) corresponds to where the function \(f(x)\) is strictly increasing.

To delve deeper into what it means for a function to be 'strictly increasing,' we refer to the behavior where, as \(x\) gets larger, \(f(x)\) does as well—without exception. This is different from a function that is simply 'increasing,' which may remain constant over some intervals.

In terms of calculus, a strictly increasing function must have a positive first derivative. The first derivative reflects the slope of the tangent line to the function at any given point; a positive slope thus means an uphill movement from left to right.

The condition \( f'(x) > 0 \) is crucial. For the function \( \tan^{-1}(\sin x + \cos x) \) in the exercise, we're seeking intervals where \( f'(x) \) maintains a strictly positive value, which led to the interval \( [0, (\pi / 4)] \) as the correct answer. Such intervals denote the domains over which the function's output consistently increases with its input.

The inverse trigonometric functions allow us to find angles when we know the ratios of the sides in a right triangle. Among these functions is \( \tan^{-1} \), also known as arctan, which is the inverse of the tangent function.

In the context of our problem, \(f(x) = \tan^{-1}(\sin x + \cos x)\) involves the arctan of a sum involving sine and cosine functions. These types of functions can be complex to differentiate and understand without trigonometric identities.

When dealing with \( \tan^{-1} \) as part of a function to be differentiated, we apply the chain rule and the derivative of \( \tan^{-1} \) combined with other functions inside it. It's the understanding of these principles that allows us to manipulate and differentiate the function effectively to find where it is increasing, as demonstrated in solving the given problem.