The function provided is:
\[f(x) = \tan^{-1} (\sin x + \cos x).\]
Step 1: Compute the derivative of the function.
The derivative of \( f(x) \) is:
\[f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot (\cos x - \sin x).\]
Simplifying using the trigonometric identity \( \sin x + \cos x = \sqrt{2} \sin \left( x + \frac{\pi}{4} \right) \):
\[f'(x) = \frac{\sqrt{2} \cos \left( x + \frac{\pi}{4} \right)}{1 + (\sin x + \cos x)^2}.\]
Step 2: Determine the interval where \( f'(x)>0 \).
For \( f'(x)>0 \), the numerator \( \sqrt{2} \cos \left( x + \frac{\pi}{4} \right) \) must be positive, which implies:
\[\cos \left( x + \frac{\pi}{4} \right)>0.\]
The cosine function is positive in the first and fourth quadrants. Solving the inequality:
\[-\frac{\pi}{2}<x + \frac{\pi}{4}<0.\]
Rearranging to solve for \( x \):
\[-\frac{\pi}{2} - \frac{\pi}{4}<x<0 + \frac{\pi}{4}.\]
This simplifies to:
\[-\frac{3\pi}{4}<x<\frac{\pi}{4}.\]
Step 3: Confirm the increasing behavior of the function.
Within the interval \( \left( -\frac{3\pi}{4}, \frac{\pi}{4} \right) \), \( f'(x)>0 \), indicating that \( f(x) \) is an increasing function.
Final Answer:
\[\boxed{\left( -\frac{3\pi}{4}, \frac{\pi}{4} \right)}\]