Tangent line using trigonometric derivatives

Differentiate term by term: $f'(x) = \cos(x) - \sin(x)$, using $\frac{d}{dx}[\sin x] = \cos x$ and $\frac{d}{dx}[\cos x] = -\sin x$. Next, evaluate the slope at $x = \pi/4$: $f'(\pi/4) = \cos(\pi/4) - \sin(\pi/4) = \frac{\sqrt2}{2} - \frac{\sqrt2}{2} = 0$. So the tangent line is horizontal at this point.

Now find the point of tangency: $f(\pi/4) = \sin(\pi/4) + \cos(\pi/4) = \frac{\sqrt2}{2} + \frac{\sqrt2}{2} = \sqrt2$. Using the point-slope form $y - f(\pi/4) = f'(\pi/4)(x - \pi/4)$ with slope $0$, the tangent line simplifies to $y = \sqrt2$.