Find y' for y = sin x + cos x
Find $y'$ for $y = \sin x + \cos x$.
Explanation of Method
Differentiate a sum of the two basic trigonometric functions.
Step 1: Differentiate term by term
Using the standard derivatives $(\sin x)' = \cos x$ and $(\cos x)' = -\sin x$: $y' = \frac{d}{dx}(\sin x) + \frac{d}{dx}(\cos x) = \cos x + (-\sin x)$.
Step 2: Verification
\[x = \frac{\pi}{4}:\quad y' = \cos\frac{\pi}{4} - \sin\frac{\pi}{4} = \frac{\sqrt2}{2} - \frac{\sqrt2}{2} = 0 \;\;\checkmark\]
\[y = \sqrt{2}\,\sin\!\left(x+\frac{\pi}{4}\right) \;\Longrightarrow\; y' = \sqrt{2}\,\cos\!\left(x+\frac{\pi}{4}\right) = \cos x - \sin x \;\;\checkmark\]
The vanishing derivative at $x = \tfrac{\pi}{4}$ matches the fact that $y$ attains its maximum value $\sqrt{2}$ there.
Result
Final answer:
\[\boxed{\,y' = \cos x - \sin x\,}\]
Remark
Remark. Differentiating four times returns the original function, since each differentiation advances the phase by $\tfrac{\pi}{2}$: $y'''' = y$. The same holds for $\sin x$ and $\cos x$ separately.