To establish that the function \( f(x) = \tan^{-1}(\sin x + \cos x) \) is increasing on the interval \( \left[ 0, \frac{\pi}{4} \right] \), we must demonstrate that its derivative, \( f'(x) \), is positive over this interval. The differentiation of \( f(x) \) with respect to \( x \) proceeds as follows, utilizing the chain rule: \[ f'(x) = \frac{d}{dx} \left( \tan^{-1}(\sin x + \cos x) \right) \] The derivative of \( \tan^{-1}(u) \) with respect to \( u \) is \( \frac{1}{1 + u^2} \). Applying this and the chain rule yields: \[ f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot \frac{d}{dx} (\sin x + \cos x) \] Differentiating \( \sin x + \cos x \) gives: \[ \frac{d}{dx} (\sin x + \cos x) = \cos x - \sin x \] Consequently, the derivative of \( f(x) \) is: \[ f'(x) = \frac{\cos x - \sin x}{1 + (\sin x + \cos x)^2} \] We now evaluate the sign of \( f'(x) \) for \( x \in \left[ 0, \frac{\pi}{4} \right] \). Within this interval, \( \cos x - \sin x \) is positive because \( \cos x>\sin x \). The denominator, \( 1 + (\sin x + \cos x)^2 \), is always positive. Therefore, \( f'(x)>0 \) on \( \left[ 0, \frac{\pi}{4} \right] \), confirming that \( f(x) \) is an increasing function on this interval.